Consider the following category of exercises: you're given two cyclic homeomorphism groups, say $G=\langle g \rangle $, $H= \langle h \rangle$, acting on $\mathbb R^2$; you're asked to determine if $\mathbb R^2 /G $ and $\mathbb R^2 /H $ are coverings of $\mathbb R^2/ \langle g,h \rangle $.
Now, it is natural to ask when $(\mathbb R^2 /G)/H=( \mathbb R^2 /H)/G $, because if this condition holds one can construct $\mathbb R^2 /G$ (for example) and verify that $H $ has a properly discontinuous action on it. It seems to me that this is the case when the group $\langle g,h \rangle $ is commutative, because that means that if two points in $\mathbb R^2$ glued by the action of $G $, then their images through an element of $H $ are still two points glued by the action of some element in $ G $. However I don't know if there is a weaker sufficient condition that allows to resolve these exercises in this way. Any clarify or information is welcome, thanks in advance.