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I'm reading Berlekamp/Conway/Guy's Winning Ways for Your Mathematical Plays. Here:

I am a little bit confused: What is happening here? It seems to me that we know that a game with a unique red edge is a $1-$move advantage for red. But we still can't know what is the advantage value for $(a)$, so we call the advantage of red and blue $r,b$. Then for $(a)$, we have $r,b$ advantages.

For $(b)$, we have $r+1,b-1$ advantages. Now $(c)$ is a zero position, it seems this allow us to write the following advantage equations: $2r+1=0, 2b-1=0$ and from this we can know the advantage value of a certain game for each player.

Is my interpretation correct? I am asking what is the "moral of the story", it seems that whenever we don't know the value of a game, we can try to "compose it" with some other games (such as the game with a single red or blue edge which we know it's value) until it forms a zero position, from which we can write a system of equations, solve and find the advantage value of each player in our unknown game.

Your idea is correct. From a broader perspective, the set of (red-blue) Hackenbush positions (up to equivalence) form a totally ordered abelian group (called the surreal numbers): they have operations of addition and subtraction and a relation $leq$ which satisfy all the usual properties. Now, it's a theorem that any totally ordered abelian group $G$ satisfying a certain extra "finiteness" condition (the Archimedean axiom) is isomorphic to a subgroup of the real numbers. Namely, if you fix some element of $G$ to call "$1$", for every other element $gin G$ you can consider the set of rational numbers $frac{m}{n}$ such that $mcdot gleq ncdot 1$. This set forms a Dedekind cut in the rational numbers and so determines a real number. It can then be shown that mapping $g$ to this real number is an isomorphism of ordered abelian groups from $G$ to a subgroup of $mathbb{R}$.

Now, in the case of Hackenbush, the set of finite Hackenbush positions (up to equivalence) satisfies the Archimedean axiom, and so this theorem applies. That means that when we identify some element to be $1$, there is a canonical way to identify such positions as real numbers. We choose to let "$1$" be a position with $1$-move advantage for Left, so that we can loosely think of the number associate to a position as "the number of moves that Left is ahead by". It turns out then that the subgroup of the real numbers corresponding to finite Hackenbush positions is the group of dyadic rationals.

A Hackenbush game has a value that is a number, so you just need one number for the value of the position, not $r,b$ separately. Yes, one way to value a position is to compose it with known positions and find a combination that has $0$ value, then use algebra to determine the value of the unknown position. If we let the value of position $a$ be $v$, the value of position $c$ is $2v-1$. Once we prove that is $0$ we can find $v=frac 12$ by algebra.

Another way is to look at the options in a position. The red above blue position is ${0|1}$ because blue can move to $0$ and red can move to $1$. There is a theorem coming that the value of ${a|b}$ is the simplest number that fits between $a$ and $b$. For ${0|1}$ that is $frac 12$