Let $\Gamma_1,\Gamma_2 \subset \text{SL}(2,\mathbb{R})$ be two Fuchsian groups. Assume that they are commensurable as subgroups of $\text{GL}(2,\mathbb{R})$, that is, there exists $g \in \text{GL}2,\mathbb{R})$ such that the intersection $$\Gamma_1 \cap (g \Gamma_2 g^{-1})$$ has finite index in both $\Gamma_1$ and $g\Gamma_2g^{-1}$. Does it follow that the manifolds $\mathbb{H}^2/\Gamma_1$ and $\mathbb{H^2}/\Gamma_2$ have a common finite isometric covering space?
I think this is true, but I read that $\mathbb{H}^2/\Gamma_1$ and $\mathbb{H^2}/\Gamma_2$ having a common finite isometric covering space is equivalent to $\Gamma_1$ and $\Gamma_2$ being commensurable as subgroups of $\text{SL}(2,\mathbb{R})$ - which should be a weaker property - and this makes me think I missed something.
Your question ignores subtleties arising from the presence of torsion elements in the groups $\Gamma_1,\Gamma_2$.
Let me first assume that $\Gamma_1,\Gamma_2$ are torsion free. In that case, what you are saying is absolutely correct:
When torsion enters the scene, then you have to be a lot more careful with your category. When a Fuchsian group $\Gamma \subset PSL(2,\mathbb R)$ has torsion, the quotient $\mathbb H^2 / \Gamma$ should be thought of as an orbifold, not as a manifold. Orbifolds form a category unto their own. In this category, there are "covering" spaces, analogous to but not quite the same as covering spaces in the topological category. You can also discuss what it means for covering maps in the orbifold category to be isometric, but the meaning is not quite the same as in the topological category.
Once those issues have been cleared up by a careful study of orbifolds, then you can resume the discussion, and what you are saying can again be worded with absolutely correctness:
Omit the word orbifold from that sentence, however, and it can be false if $\Gamma_1$ or $\Gamma_2$ has torsion.