Haskell, 77 71 62 bytes

f@(d:e)#(p:q)=f:[e,f++take d q]!!p#q

_#_=

a!b=tail$a#cycle b

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Edit: -9 bytes thanks to @xnor.

C# (Visual C# Interactive Compiler), 82 bytes

m=>n=>{for(int i=0;m!="";Print(m=n[i++]<49?m.Substring(1):m[0]>48?m+n[i]:m))n+=n;}

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J, 65 bytes

(([:(][echo)(}.@[)`([,{.@[#1{],])@.({.@]));1|.])&>/^:(0<0#@{>)^:5

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I may golf this further later. Note the 5 at the end would be infinity _ in the actual program, but I've left it there to make running the non-halting examples easier.

Python 3, 74 bytes

def f(d,p):

 while d:c,*p=p+p[:1];d=(d[1:],d+p[:1]*d[0])[c];d and print(d)

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Arguments: d: data, p: program.

05AB1E, 24 21 bytes

[¹Nèi¬i¹N>è«}ë¦}DõQ#=

Takes the program as first input and data as second input.input.

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Explanation:

[             # Start an infinite loop:

 ¹Nè          #  Get the N'th digit of the first (program) input

              #  (NOTES: N is the index of the infinite loop;

              #          indexing in 05AB1E automatically wraps around)

    i         #  If this digit is 1:

     ¬        #   Push the head of the current data (without popping it)

              #   (will take the second (data) input implicitly if it's the first iteration)

      i     } #   If this head is 1:

       ¹N>è   #    Get the (N+1)'th digit of the first (program) input

           «  #    And append it to the current data

    ë }       #  Else (the digit is a 0 instead):

     ¦        #   Remove the first digit from the current data

              #   (will take the second input (data) implicitly if it's the first iteration)

 DõQ          #  If the current data is an empty string:

    #         #   Stop the infinite loop

 =            #  Print the current data with trailing newline (without popping it)

Ruby, 62 59 bytes

->c,d{p(d)while(a,*c=c;b,*d=d;c<<a;!=d=[b]*a+d+c[0,a*b])}

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How

• Get the first bit from code c and data d, call them a and b. Put a back at the end of c.


• Put back b at the beginning of d if a==1. This can be shortened to [b]*a
  


• Put the first byte of c at the end of d if a==1 and b==1. This can be shortened to c[0,a*b].


• If we have more data, print and repeat.

Python 2, 96 82 bytes

def g(d,p):

 while d:

	c=p[0];p=p[1:]+[c];d=[d[1:],d+[p[0]]*d[0]][c]

	if d:yield d

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Stealing a bit from Emodiment of Ignorance's answer...

A generator which uses lists of 1's and 0's for input / output.

Jelly, 40 bytes

;€Ø2œịxØ1œị$Ʋ$Ḋ€2,1œị$?1¦ṙ€1$2¦µ⁺1ịGṄƲ¿Ḣ

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I’ve assumed trailing newlines are ok. I’ve also gone with a list of two lists of zeros and ones as input, and output to stdout.

Python 1, 75 bytes

a,b=input()

while a:b=b[1:]+b[:1];a=[a[1:],a+b[:1]*a[0]][b[0]];print a or''

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C++ (gcc), 178 bytes

void a(std::string s,std::string d){while(!s.empty())for(int i=0;i<d.size();i++){if(d[i]=='0')s.erase(0,1);else if(s[0]=='1')s.push_back(d[(i+1)>=d.size()?0:i+1]);std::cout<<s;}}

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C++ (gcc), 294 289 bytes

-5 bytes thanks to @ceilingcat

#import<iostream>

#import<queue>

void a(char*e,char*p){std::queue<char>d;for(;*e;)d.push(*e++);for(char*c=p;d.size();c=*++c?c:p){*c-49?d.pop():d.front()-48?d.push(c[1]?c[1]:*p):a("","");if(d.size()){for(int i=0;i++<d.size();d.pop())std::cout<<d.front(),d.push(d.front());std::cout<<" ";}}}

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Fairly straightforward algorithm. Copies the data into a queue, and repeatedly loops through the program. On a "0", it removes the first element in the queue (the first "bit"). On a 1, it adds the next "bit" of the program to the data if the first "bit" of the data is 1. Then it loops through the data, printing it "bit" by "bit", and finally prints a space to separate successive data entries.