Suppose that $M$ is a smooth manifold, and $G$ is a subgroup of homeomorphisms which acts naturally on the points of $M$. Suppose also, that the action is properly discontinuous, i.e. for any point $p$, we can find an open neighborhood $U_p$ of $p$ such that for any $g \in G\setminus \{Id\}$, $g(U_p) \cap U_p = \emptyset $.
Then the restriction of the projection $\pi : M \rightarrow M/G$ to $U_p$ becomes a homeomorphism.
I am not arguing with that, but what I cannot comprehend is how does the inverse of $\pi \vert _{U_p} $ work? How does it choose which representative of the class to send back?
Is it because there is a unique $q_0$ which belongs to $U_p$, for any class in $\pi (U_p)$? Can we, for some intents and purposes, assume that $\pi\vert_{U_p}^{-1}([q])=q$? For example, when using it to define charts for $M/G$, and we want to prove smoothness for some certain function of $M/G$?
Here is ho to construct the inverse to $\pi|_{U_p}$: Elements of $M/G$ are $G$-orbits $Gx, x\in M$. Given such $Gx$, take $f(Gx)=Gx\cap U_p$. This intersection could be empty, but when nonempty it consists of exactly one point. I will leave it to you to check that the intersection is nonempty if and only if $Gx\in \pi(U_p)$, that $f|_{\pi(U_p)}$ is inverse to $\pi|_{U_p}$ and is continuous.