I would do this computation by thinking about covering spaces of graphs and applying Stallings's *folding* algorithm.

We realise $F_2$ as the fundamental group of a graph $X$ with one vertex $v$ and two edges---this is sometimes called *the rose with two petals*. We orient the edges, and label one by $a$ and the other by $b$. This fixes an identification $\langle a,b\rangle\equiv\pi_1(X,v)$.

Your subgroup $H$ can be thought of similarly. It is the fundamental group of a graph $Y$, which we construct as follows:

- Fix a base vertex $*$.
- Attach both ends of an oriented edge labelled $a$ to $*$ .
- Attach both ends of an oriented interval, consisting of $2k$ edges labelled $a$ and $b$ alternately, to $*$.

The orientations and labels define a natural map $Y\to X$, and the image of $Y$ is your subgroup $H$.

(For more on this sort of construction see, for instance, this blog post.)

Stallings' folding algorithm is a way of turning this map into an immersion---that is, a local embedding. The algorithm is easy:

- If two edges with the same label are both oriented into the same vertex, identify them.
- If two edges with the same label are both oriented away from the same vertex, identify them.
- Repeat.

At the end of this procedure, we have a new oriented, labelled graph $Y'$, and the map $Y'\to X$ is an immersion. There are essentially two possibilites:

Every vertex of $Y'$ has valence four. If so, then $Y'\to X$ is a covering map and $H$ is a finite-index subgroup of $F_2$. The index of $H$ is equal to the degree of the covering map, which is equal to the number of vertices of $Y'$.

Some vertex of $Y'$ has valence less than four. If so, then $H$ is of infinite index. (To see this, note that you can complete it to an infinite-sheeted covering space.)

In your case, you can quickly see that you need to perform exactly one fold to turn $Y$ into $Y'$. If $k=1$ then $Y'$ is isomorphic to $X$ and your subgroup $H$ is equal to $F_2$. Otherwise, $Y'$ has a vertex of valence two and $H$ is of infinite index.

~~has index 3 in SL(2,Z) =~~– KCd Jun 20 '11 at 20:49~~, for any k > 2 the subgroup~~~~has infinite index in SL(2,Z). The proof, shown to me by Vincentiu Pasol, uses the action of SL(2,Z) on primitive vectors in Z^2. Because of this result,[F:H] is infinite for k > 2.~~