If we have G acts properly discontinuously on hyperbolic plane $\mathbb H$, then for any point p $\in \mathbb H$, exist neighborhood V s.t. gV$\cap$V =$\emptyset$ iff gp$\neq$p.
Given this, can we find V s.t. $\pi$(V) is homeomorphic to V? Here $\pi$ is the natural map from $\mathbb H$ to G/$\mathbb H$, i.e. send p to the orbit containing p.
If I understand your question correctly:
Notice that $h(gV)=g(hV)$ and so $g(hV)\cap hV\ne\emptyset$. So using one neighborhood you find one for all points. Basically this gives what you want, although my personal preference would be to generate the topology using a fundamental domain.