On which structures does the free group 'naturally' act?

$F_2$ acts on a certain tiling of the hyperbolic plane. It looks sort of like this:

The above tiling is acted on by the modular group $\Gamma \cong \text{PSL}_2(\mathbb{Z})$, which naturally sits as a subgroup inside of the full group $\text{PSL}_2(\mathbb{R})$ of isometries of the hyperbolic plane. Abstractly, this group is the free product $C_2 \ast C_3$. It has congruence subgroups $\Gamma(N)$ given by the image of the kernel of the quotient maps $\text{SL}_2(\mathbb{Z}) \to \text{SL}_2(\mathbb{Z}/N\mathbb{Z})$ in $\Gamma$. For $N \ge 2$ these groups act freely on the hyperbolic plane, and the topological structure of their quotients are known (they are uncompactified modular curves, all of which will be compact Riemann surfaces minus finitely many points, and both the genus and the number of points removed are known). In particular, $\mathbb{H}/\Gamma(2)$ is known to be $\mathbb{R}^2$ minus two points, which has fundamental group $F_2$. Hence

$$\Gamma(2) \cong F_2$$

acts on the hyperbolic plane $\mathbb{H}$. You can find a fundamental domain for this action by taking a union of finitely many fundamental domains for the action of $\Gamma$ (pictured above), and this gives a tiling of the hyperbolic plane whose automorphism group should be exactly $\Gamma(2)$ (I think).

Groups that arise in this way can be thought of as hyperbolic analogues of wallpaper groups.

In general, an interesting way to find an interesting space on which a group $G$ acts is to find a space $X$ with $\pi_1(X) \cong G$ and look at the action of $G$ on the universal cover $\tilde{X}$, especially if $X$ has extra structure for its universal cover to inherit. In the case of $F_2$ we can pick $X$ to be the wedge of two circles and this reproduces the action of $F_2$ on its Cayley graph, but for other kinds of spaces $X$ we get more interesting spaces. Above we instead pick $X$ to be $\mathbb{R}^2$ minus two points and get an action on the hyperbolic plane; this is a special case of uniformization.

Edit: Perhaps an easier description of what I'm talking about is in terms of the von Dyck groups. In the Poincaré disk model the relevant tiling looks like this:

$F_2$ is (Edit:) the group of orientation-preserving symmetries of this tiling that can be obtained by rotating around the vertices of any of the triangles in it. (The original statement that was here was wrong; the tiling above clearly admits a rotational symmetry of order $3$, but $F_2$ has no torsion. The resolution is that this rotational symmetry fixes the origin, which is not a vertex.)