JavaScript (ES6),  130  126 bytes

Takes input as (letters)(words). Returns $0$ or $1$.

L=>W=>L.every(a=>a.every(x=>(W+'').match(x,a.map(y=>s+='|'+x+y))),p=s=1)&W.every(w=>w[2]&&p|w[0]==p&!w.match(s,p=w.slice(-1)))

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Step 1

We first iterate over $L$ to build a pipe-separated string $s$ consisting of all invalid pairs of letters. While doing so, we also make sure that each letter appears at least once in some word.

L.every(a =>              // for each group of letter a in L:

  a.every(x =>            //   for each letter x in a:

    (W + '')              //     coerce W to a string

    .match(               //     and test whether ...

      x,                  //       ... x can be found in it

      a.map(y =>          //       for each letter y in a:

        s += '|' + x + y  //         append '|' + x + y to s

      )                   //       end of map()

    )                     //     end of match()

  ),                      //   end of inner every()

  p = s = 1               //   start with p = s = 1

)                         // end of outer every()

Step 2

We now iterate over $W$ to test each word.

W.every(w =>              // for each word w in W:

  w[2] &&                 //   is this word at least 3 characters long?

  p |                     //   is it the first word? (p = 1)

  w[0] == p &             //   or does it start with the last letter of the previous word?

  !w.match(               //   and finally make sure that ...

    s,                    //     ... it doesn't contain any invalid pair of letters

    p = w.slice(-1)       //     and update p to the last letter of w

  )                       //   end of match()

)                         // end of every()

Python 2, 171 bytes

lambda l,w:(set(sum(l,))==set(''.join(w)))*all(a[-1]==b[0]for a,b in zip(w,w[1:]))*all((a in g)+(b in g)<2for x in w for a,b in zip(x,x[1:])for g in l)*min(map(len,w))>2

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05AB1E, 37 35 33 32 bytes

εSk3÷üÊ}¹ü)ε`нsθQ}²{¹SêQ¹€g3@)˜P

-2 bytes by taking inspiration of the ê approach @Emigna used in his 05AB1E answer.

Takes the string of words as first input, and the flattened list of twelve letters as second input.

Try it online or verify all test cases.

Explanation:

ε      }    # Map over the strings `y` of the (implicit) input-list of words:

 S          #  Convert word `y` to a list of characters

  k         #  Get the index of each character in the (implicit) input-list of letters

   3÷       #  Integer-divide each index by 3

     üÊ     #  Check for each pair of integers that they are NOT equal

¹ü)         # Get all pairs of strings from the input-list of words

   ε     }  # Map over the pair of words:

    `       #  Push both to the stack

     н      #  Leave only the first character of the second word

      sθ    #  And the last character of the first word

        Q   #  And check if they are equal

²{          # Push the input-list of letters, and sort them

  ¹S        # Push the input-list of words, converted to characters and flattened

    ê       # And then uniquified and sorted as well

     Q      # And check if both lists of characters are the same

¹€g         # Get the length of each string in the input-list of words

   3@       # Check for each length if it's >= 3

)˜          # Then wrap everything on the stack into a list, and deep flatten it

  P         # And check if everything is truthy by taking the product

            # (which is output implicitly as result)

05AB1E, 42 bytes

εg2›}P¹εεUIεXå}ƶO}üÊP}P¹ü‚ε`нsθQ}P¹Jê²JêQP

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Jelly, 34 bytes

Ṫ=¥/ƝḢ€Ạȧ⁸Fe€ⱮZḄ;IẠƊȧF}fƑF{

ẈṂ>2ȧç

A dyadic Link accepting the words on the left and the letter groups on the right which yields 1 if valid and 0 if not.

Try it online! Or see the test-suite.

Haskell, 231 bytes

import Data.List

l&w=all((>2).length)w&&c w&&all(l!)w&&(h l)%(h w)

h=concat

l%w=null[x|x<-l,x`notElem`w]

l!(a:b:c)=a#l?(b#l)&&l!(b:c)

l!_=1>0

Just a?Just b=a/=b

_?_=1<0

c#l=findIndex(elem c)l

c(a:b:t)=last a==head b&&c(b:t)

c _=1>0

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Not the best score. Some Haskell guru will probably be able to get this under 100 bytes.

Usage

["ICO","MRE","GNS","APL"]&["CROPS", "SAIL", "LEAN", "NOPE", "ENIGMA"]

Explanation

import Data.List

l&w = all((>2).length)w &&      -- Every word has length > 2

      c w &&                    -- Every word ends with the same letter as the next one starts with

      all(l!)w &&               -- For every word: Consecutive letters are on different sides (and must exist on a side)

      (h l)%(h w)               -- All letters are used



h=concat                        -- Just a shorthand



l%w=null[x|x<-l,x`notElem`w]    -- The letters of l, with all letters of w removed, is empty



l!(a:b:c)=a#l?(b#l)&&l!(b:c)    -- Sides of the first two letters are different, recurse from second letter

l!_=1>0                         -- Until fewer than 2 letters remain



Just a?Just b=a/=b              -- Both sides must be different

_?_=1<0                         -- And must exist



c#l=findIndex(elem c)l          -- Find the side of letter c



c(a:b:t)=last a==head b&&c(b:t) -- Last letter of the first word must be same as first letter of second word, recurse starting from second word

c _=1>0                         -- Until there are fewer than 2 words

Jelly, 30 bytes

FQṢ=Ṣ},i@€€’:3Iʋ,Ẉ>2ɗ,Ṫ=Ḣ}ɗƝ{Ȧ

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A dyadic link that takes the list of words as left argument and the flattened list of letters in the box as the right argument. It returns 1 for true and 0 for false.

Explanation

F                               | Flatten the word list

 Q                              | Unique

  Ṣ                             | Sort

   =                            | Is equal to

    Ṣ}                          |   The sorted letterbox letters

      ,        ʋ                | Pair this with the following:

       i@€€                     |   The index of each letter of each word in the letterbox            

           ’                    |   Decrease by 1

            :3                  |   Integer divide by 3

              I                 |   Differences between consecutive ones (will be zero if any two consecutive letters in a word from same side of box)

                ,   ɗ           | Pair everything so far with the following:

                 Ẉ>2            |   Whether length of each input word is greater than 2

                     ,    ɗƝ{   | Pair everything so far with the following, applied to each neighbouring pair of the input word list

                      Ṫ         |   Last character of first word

                       =Ḣ}      |   Equals first character of second

                             Ȧ  | Flatten all of the above and check there are no false values

Haskell, 231 bytes

A different Haskell variation, exactly the same size as @Paul Mutser's :)

import Data.List

f x=filter(a->length a>1)$concatMap subsequences x

g=nub.concat.f

p l(x:y)=foldl((m,n)c->(c,n&&length c>2&&(not$any(`isInfixOf`c)(f l))&&last m==head c))(x,True)y

z l w=null(g l\g w)&&null(g w\g l)&&(snd$p l w)

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Ungolfed

-- generate all invalid substrings

f :: [String] -> [String] 

f xs = filter (x -> length x > 1) $ concatMap subsequences xs



-- utility function to flatten and remove duplicates

g :: [String] -> String

g  = nub $ concat $ f



-- verify that all conditions are satisfied along the list

p :: [String] -> [String] -> (String, Bool)

p l (x:xs) = foldl ((m,n) c -> (c , n && length c > 2 && (not $ any (`isInfixOf` c)(f l)) && last m == head c)) (x, True) xs



-- put all the pieces together and consume input

z :: [String] -> [String] -> Bool

z l w = null (g l \ g w) && null (g w \ g l) && (snd $ p l w)

Python 2, 168 156 bytes

lambda l,w,J=''.join:(set(J(w))==set(J(l)))*all((v<1or u[-1]==v[0])*u[2:]*(2>(x in p)+(y in p))for u,v in zip(w,w[1:]+[0])for x,y in zip(u,u[1:])for p in l)

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Returns 1 for truthy, 0 for falsey.

Charcoal, 63 bytes

⌊⁺⁺⁺⭆η›Ｌι²⭆⪫ηω№⪫θωι⭆⪫θω№⪫ηωι⭆η⭆ι⎇μ¬⁼Φθ№νλΦθ№ν§ι⊖μ∨¬κ⁼§ι⁰§§η⊖κ±¹

Try it online! Link is to verbose version of code. Explanation:

⌊⁺⁺⁺

Concatenate the below expressions and output 0 if any of them includes a 0 otherwise 1.

⭆η›Ｌι²

For each word in the solution output whether its length is at least 3.

⭆⪫ηω№⪫θωι

For each letter in the solution output whether it appears in the puzzle.

⭆⪫θω№⪫ηωι

For each letter in the puzzle output whether it appears in the solution.

⭆η⭆ι⎇μ¬⁼Φθ№νλΦθ№ν§ι⊖μ∨¬κ⁼§ι⁰§§η⊖κ±¹

For each letter in the solution check that the previous letter is not in the same group, unless it is the first letter of a word, in which case check that it is equal to the last letter of the previous word, unless it is the first letter of the solution, in which case just ignore it.

Java (JDK), 190 bytes

g->w->{var v=0<1;int x=0,l,i=0,j,p,z,y=w[0][0];for(;i<w.length;i++)for(l=w[i].length,v&=y==w[i][0]&l>2,j=0,p=-9;j<l;v&=z>=0&z/3!=p/3,x|=1<<1+(p=z))z=g.indexOf(y=w[i][j++]);return v&x==8190;}

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Explanations

g->w->{     // Lambda accepting letter groups as a string and a list of words, in the form of an array of char arrays.

 var v=0<1;     // Validity variable

 int x=0,       // The letter coverage (rule 4)

     l,         // The length of w[i]

     i=0,       // The w iterator

     j,         // The w[i] iterator

     p,         // The previous group

     z,         // The current group

     y=w[0][0]; // The previous character

 for(;i<w.length;i++) // For each word...

  for(

     l=w[i].length,     // make a shortcut for the length

     v&=y==w[i][0]&l>2, // check if the last character of the previous word is the same as the first of the current.

                        // Also, check if the length is at least 3

     j=0,               // Reset the iteration

     p=-9               // Set p to an impossible value.

    ;

     j<l                // 

    ;

     v&=z>=0&z/3!=p/3,  // Check that each letter of the word is in the letter pool,

                        //  and that the current letter group isn't the same as the previous one.

     x|=1<<1+(p=z)      // After the checks, assign z to p,

                        //  and mark the letter of the pool as used.

   )

   z=g.indexOf(y=w[i][j++]); // Assign the current letter to y so that it contains the last at the end of the loop.

                             //  and fetch the position of the letter in the pool.

 return v&x==8190; // Return true if all matched

                   //  and if the rule 4 is enforced.

}