I am thinking that: If we can find a compact set $K \subset SL(2,\mathbb{R})$ such that $(\pi_3 \circ \pi_2 \circ \phi \circ \pi_1 )(K)=\Gamma \backslash \mathbb{H}$ then $\Gamma \backslash \mathbb{H}$ is compact (by continuity of maps $\pi_3,\pi_2,\phi,\pi_1$, a compact set maps to a compact set). That is equivalent to find a fixed compact set $K \subset SL(2,\mathbb{R})$ such that for every $g\in SL(2,\mathbb{R})$ there exist $\gamma \in \Gamma^1$ such that $\gamma.g\in K$.
If you find a compact subset $K \subset \text{SL}(2,\mathbb{R})$ so that the map from $K$ to $\Gamma \setminus \mathbb{H}$ is surjective, then you can find another compact subset $K' \subset \text{SL}(2,\mathbb{R})$ so that the second condition holds. To construct $K'$, you will need the fact that every point pre-image of the map $\text{SL}(2,\mathbb{R}) \mapsto \Gamma \setminus \mathbb{H}$ is compact, being homeomorphic to $S^1 \times \text{(two points)}$. I will refer to these point pre-images as "fibers". Then you simply take $K'$ to be the union of $K$ with each fiber that has nonempty intersection with $K$.