For the rest of this post, let
$$ A=\begin{pmatrix} 1 & 0\\ 2 & 1\end{pmatrix}, B=\begin{pmatrix} 1 & 2\\ 0 & 1\end{pmatrix}, C=\begin{pmatrix} -1 & 0\\ 0 & -1\end{pmatrix}, D=\begin{pmatrix} -1 & 0\\ 0 & 1\end{pmatrix}.$$
Note that all of $A$, $B$, $C$ and $D$ live in $\Gamma(2)$.
First, let's consider the case of $G=SL(2,\mathbb{Z}$), which is the case I think you wanted (so in your notation, we have the requirement that $ad-bc=1$). Note that in this case, $D\notin G$.
Proposition: $A$, $B$, and $C$ generate $\Gamma(2)$.
Proof: Define a mapping from $f:\ \Gamma(2)\rightarrow \mathbb{Z}^+$ by the formula
$$ f:\ \begin{pmatrix} a & b\\ c & d\end{pmatrix}\mapsto |a|+|c|.$$
Let $\mathfrak{H}$ be the subgroup of $\Gamma(2)$ generated by $A$, $B$, and $C$, and let $X$ be an arbitrary element of $\Gamma(2)$. We will be done if we can show $X\in \mathfrak{H}$.
To this end, pick an element $Y\in \mathfrak{H}X$ [the right coset of $\mathfrak{H}$ containing $X$] for which $f(Y)$ is minimal.
Now letting $Y=\begin{pmatrix} a & b\\ c & d\end{pmatrix}$, consider the following cases:
- $c=0$.
We know $ad-bc=1$, and so in this case $a=d=\pm 1$. But then $Y$ (or $YC$) must be a power of $B$, since
$$ B^n=\begin{pmatrix} 1 & 2n\\ 0 & 1\end{pmatrix}.$$
This means $Y\in\mathfrak{H}\cap\mathfrak{H}X$, so that $\mathfrak{H}=\mathfrak{H}X$, or $X\in \mathfrak{H}$.
- $c\neq0$, and $|a| > |c|$.
Then there exists an $n\in\mathbb{Z}$ such that $-|c| < a+2nc < |c|$ [strict inequality because $a$ is odd and $c$ is even], and then
$$ B^nY=\begin{pmatrix} 1 & 2n\\ 0 & 1\end{pmatrix}\begin{pmatrix} a & b\\ c & d\end{pmatrix}=\begin{pmatrix} a+2nc & b+2nd\\ c & d\end{pmatrix},$$
so that $f(B^nY)=|a+2nc| + |c| < |c| + |c| < |a| + |c|$, contradicting the choice of $Y$. In other words, this case does not happen.
- $c\neq 0$, and $|a| < |c|$. Then a similar argument to the one above, using $A$ instead of $B$, leads also to a contradiction. The proof is now complete.
Thus we see that if $Z=\langle C\rangle$, then $\Gamma(2)/Z$ is generated by $A$ and $B$. But the group generated by $A$ and $B$ is free, by the Ping-Pong Lemma.
I think this is enough for the question, but if you actually meant $G=GL(2,\mathbb{Z})$, one can still say a lot. In this case, $\Gamma(2)/Z$ is not free, but has $F_2$ as a subgroup of index 2. There are only a handful of groups which possess $F_2$ as a subgroup of index 2, and in this case one gets the isomorphism $\Gamma(2)/Z\cong F_2\rtimes C_2$, where $F_2=\langle A, B\rangle$ and $C_2=\langle D\rangle$, with $D$ acting via $A^D=A^{-1}, B^D=B^{-1}$.
To elaborate on Martin's comment: let $x_i = ab^i a^{-1}$. Then what you have said is that $x_1 x_4 = x_5$ in what you are calling $F(S)$ (what you really mean is $\langle S\rangle$, the subgroup of $F_2$ generated by $S$). But $x_1 x_4$ is a reduced word in the $x_i$'s, so it shouldn't 'collapse to' a different word. Since it does, $\langle S\rangle$ is not free on $S$.
EDIT: Since what you were mainly asking about was whether your line of reasoning is correct, I'll elaborate a little more. When you define $F(S)$ to be the free group on $S$, then you are treating $S$ as an abstract set, rather than as a subset of $F_2$. This is perfectly valid in some sense, but it's not going to give you a subgroup of $F_2$, so it's not helpful.
What you want to do is to find, for each $k\in \mathbb{N}$, a subset $X_k$ of $F_2$ such that the subgroup generated by $X_k$ is free on $X_k$.
You're actually very close to the right idea, though. The standard way of doing this does involve taking conjugates, but remember that you don't want any 'collapsing' to occur except when you have something like $x_i x_i^{-1}$, so you'll need to do something a little differently.
Well, since there's another answer with an outline of a proof, and where to look etc., I'll add a partial spoiler for this way, but please don't look until you've tried.
Instead of setting $x_i = ab^i a^{-1}$, try $x_i = a^i b a^{-i}$. Now all that pesky 'collapsing' shouldn't be a problem. So if $S_k = \langle x_i \mid 1\leq i\leq k\rangle$, you should be able to show that $S_k\cong F_k$.
Best Answer
If $F_3$ were a quotient of $F_2$, then $\mathbb{Z}^3$ would be, but $\mathbb{Z}^3$ cannot be generated by fewer than $3$ elements. To me it seems easier to see directly that $\mathbb{Z}^3$ needs at least $3$ generators than the corresponding statement for $F_3$, perhaps because it's easy to visualize.
The rank of a group is the smallest cardinality of a generating set. Here's a list of some facts about ranks of groups (including that the rank of $F_3$ is $3$) on Wikipedia.