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It's also easy to see that these subgroups are the minimal-index torsion-free subgroups: any covering of $G$ by a genuine graph $\widehat{G}$ must have even degree, so that the vertices above $u$ have trivial label; and it must have degree dividing divisible by three, so that the vertices above $v$ have trivial label.
Exactly the same argument can be made to work for $SL_2(\mathbb{Z})$, which you of course should think of as $\mathbb{Z}/4*_{\mathbb{Z}/2}\mathbb{Z}/6$. Indeed, if I have it right, these techniques should prove the following theorem, which is presumably standard, although I don't know a reference. (Warning: I haven't thought about this very hard.)
(At the moment what I'm seeing doesn't quite square with some of the other answers/comments: I get one normal free subgroup of $PSL_2(\mathbb{Z})$ of index six, and one conjugacy class of non-normal free subgroups of index six, corresponding to a regular covering of $G$ and an irregular covering, respectively; this seems at odds with Mark's remark about different maps to $\mathbb{Z}/12$. \mathbb{Z}/12$. But it's late, so I'm not going to figure out what's going on now.) Update I had forgotten an important subtlety. A covering map of graphs of groups$\widehat{G}\to G$includes an extra piece of information, viz: For each vertex$\hat{v}$of$\widehat{G}$with image$v$in$G$, and for each edge$e$incident at$v$, a covering map specifies a bijection between the set of edges of$\widehat{G}$incident at$\hat{v}$that map to$e$and the set of double cosets$\widehat{G}_{\hat{v}}\backslash G_v/G_e$(where$G_v$is the group labelling$v$, etc). As such bijections always exist, this doesn't arise if you want to prove the existence of a subgroup. But the correct notion of equivalence for covering maps takes these bijections into account, so two different covering maps$\widehat{G}\to G$may have the same underlying graph-map. This happens for the commutator subgroup and the Sanov subgroup: in both cases, the underlying graph is the first subdivision of the theta graph. 1 I'm late to the party, but here's how I think about this fact. The approach I will present is perhaps slightly higher-tech than some, but has the advantage that it's a completely general way of computing a free finite-index subgroup of a virtually free group. To keep things simple, I'll do$PSL_2(\mathbb{Z})$, and then explain how to do$SL_2(\mathbb{Z})$at the end. As you presumably know,$\Gamma=PSL_2(\mathbb{Z})\cong \mathbb{Z}/2*\mathbb{Z}/3$. You should think of it as the fundamental group of a graph of groups$G$, with one edge and two vertices, one of which,$u$, is labelled by$\mathbb{Z}/2$and one,$v$, labelled by$\mathbb{Z}/3$. Just as in the case of ordinary topology, any (conjugacy class of a) subgroup$H\subseteq\Gamma$corresponds to a covering map$\widehat{G}\to G$. (This is a trivial consequence of Bass--Serre theory: take$\widehat{G}=T/H$where$T$is the Bass--Serre tree.) Coverings of graphs of groups are just like coverings of graphs, except that sometimes a neighbourhood looks like a quotient by a non-free action of a group, in which case you have to remember point stabilisers (cf. orbifolds). The subgroup$H$is free if and only if every vertex of$\widehat{G}$is labelled by the trivial group, in which case we can think of$\widehat{G}$as a genuine graph and$H$as its fundamental group. To prove that$PSL_2(\mathbb{Z})$has a free subgroup of index$6$, we now just have to construct a covering map$\widehat{G}\to G$of degree$6$, where$\widehat{G}$is a genuine graph. But this is easy. To do this, take three vertices$u_1,u_2,u_3$of valence two and two vertices$v_1,v_2$of valence three. Now glue them together however you like, such that each edge adjoins one$u_i$and one$v_j$. (I think there are exactly two ways of doing this, up to permuting the$i$'s and$j$'s.) The covering map$\widehat{G}\to G$sends$u_i\mapsto u$and$v_j\mapsto v$, and each edge goes to the unique edge of$G$. Whichever way you do the gluing, the graph$\widehat{G}$has rank two, so the subgroup is isomorphic to$F_2$. It's also easy to see that these subgroups are the minimal-index torsion-free subgroups: any covering of$G$by a genuine graph$\widehat{G}$must have even degree, so that the vertices above$u$have trivial label; and it must have degree dividing three, so that the vertices above$v$have trivial label. Exactly the same argument can be made to work for$SL_2(\mathbb{Z})$, which you of course should think of as$\mathbb{Z}/4*_{\mathbb{Z}/2}\mathbb{Z}/6$. Indeed, if I have it right, these techniques should prove the following theorem. (Warning: I haven't thought about this very hard.) Theorem: If$\Gamma=A*_CB$for$A,B$finite, then a minimal-index free subgroup is of index$d=\mathrm{lcm}(|A|,|B|)$, and of rank$1-d\left(\frac{1}{|A|}+\frac{1}{|B|}-\frac{1}{|C|}\right)$. Note that this technique is powerful enough to let you compute all the conjugacy classes of all the subgroups of a given index, if you want. (At the moment what I'm seeing doesn't quite square with some of the other answers/comments: I get one normal free subgroup of$PSL_2(\mathbb{Z})$of index six, and one conjugacy class of non-normal free subgroups of index six, corresponding to a regular covering of$G$and an irregular covering, respectively; this seems at odds with Mark's remark about different maps to$\mathbb{Z}/12\$. But it's late, so I'm not going to figure out what's going on now.)