If $\pi:X\to Y$ is a surjective local homeomorphism and $X$ is a smooth manifold, then does $Y$ become a smooth manifold?
This is only an answer to the original question.
Of course the minimal assumption is that $\pi$ is a surjection because $Y \setminus \pi(X)$ could be everything.
In general $Y$ need not even be Hausdorff. Let $X = \mathbb R \times \{1, 2\}$ with the obvious differentiable structure and let $Y$ be the line with two origins (call them $p_1,p_2$) which is the standard example of a "non-Hausdorff manifold" (see The Line with two origins). Define $\pi : X \to Y$ by $p(x,i) = x$ for $x \ne 0$ and $\pi(0,i) = p_i$.
So let us assume that $Y$ is Hausdorff. Since $\pi$ is a local homeomorphism, it is an open map and $Y$ is locally Euclidean. Since $X$ is a manifold, it has a countable base $\mathcal B$. It is then easy to see that $\pi(\mathcal B) = \{ \pi(B) \mid B \in \mathcal B \}$ is a (trivially countable) base for $Y$. Therefore $Y$ is a topological manifold. However, we cannot expect that there exists a differentiable structure on $Y$ such that $\pi$ is a local diffeomorphism (but note that this is a stronger requirement than $\pi$ smooth).
Let $X = \mathbb R \times \{1, 2\}$ and $Y = \mathbb R$. Define $\pi : X \to Y$ by $\pi(x,1) = x$ and $\pi(x,2) = \sqrt[3]{x}$. Next define $\pi_i : \mathbb R \to \mathbb R, \pi_i(x) = \pi(x,i)$. These maps are homeomorphisms (in fact, $\pi_1 = id$ and $\pi_2 =$ cubic root). Assume that there exists a differentiable structure $\mathcal D$ on $Y = \mathbb R$ such that $\pi$ is a local diffeomorphism. Then so are the maps $\pi_i$ and hence also $$\pi_2 = (\pi_1)^{-1} \circ \pi_2.$$ But $\pi_2$ is not even differentiable in $0$.
$\newcommand{\res}[2]{\left.#1\right|_{#2}}$ $\newcommand{\id}{{\rm id}}$ $\newcommand{\vphi}{\varphi}$ $\newcommand{\vare}{\varepsilon}$ The proof is divided into two parts.
(i) $\pi$ is a local homeomorphism. With $G$ being properly discontinuous, for any $x\in M$ we can find a neighborhood $U_0$ such that $\{g\in G:gU_0\cap U_0\neq\varnothing\}$ is a finite set. If it contains only ${\rm id}$ then we are done. If not, let the elements be $$g_1={\rm id},\ g_2,\cdots,\ g_n$$ Now by the Hausdorff property of $M$ and the fact that $G$ is free from fixed points we find nonintersecting neighborhoods $$U_1,\ \cdots,\ U_n\quad\text{of}\quad x,\ g_2x,\ \cdots,\ g_nx$$ respectively. Finally let $U_x=U_0\cap(\bigcap_{k=1}^ng_k^{-1}U_k)$. Then $U_x$ is a neighborhood of $x$ such that $g(U_x)\cap U_x=\varnothing$ for all $g\neq\id$. From this we conclude $\pi|_{U_x}:U_x\to\pi(U_x)$ is injective and hence bijective, and apparently $\pi^{-1}(U_x)=\bigcup_{g\in G}g(U_x)$ is open, it follows that $\pi|_{U_x}:U_x\to\pi(U_x)$ is a homeomorphism (the continuity of $(\pi|_{U_x})$ and $(\pi|_{U_x})^{-1}$ are easy to verify). Therefore, $\pi$ is a local homeomorphism.
(ii) $M/G$ has a structure of the same type as $M$. For each $x\in M$, from (i) there exists a neighborhood $U_x$ of $x$ such that $\res{\pi}{U_x}$ is a homeomorphism. By taking an intersection if necessary, we may assume $U_x$ is a coordinate neighborhood with the corresponding chart $\varphi_x$. We claim that the set $$\{\varphi_x(\res{\pi}{U_x})^{-1},\ x\in M\}$$ is an atlas on $M/G$. The domains of them obviously constitute an open cover of $M/G$, hence it remains to consider the transition maps, which are of the form (here, $\pi(U_x)\cap\pi(U_y)$ is assumed to be connected, as we can discuss each connected component separately) $$\vphi_x(\res{\pi}{U_x})^{-1}(\res{\pi}{U_y})\vphi_y^{-1},\quad\pi(U_x)\cap\pi(U_y)\neq\varnothing$$ It suffices to show that the middle part satisfies $$(\res{\pi}{U_x})^{-1}(\res{\pi}{U_y})=g,\quad\text{in}\quad(\res{\pi}{U_y})^{-1}(\pi(U_x)\cap\pi(U_y))$$ for some $g\in G$ because each $g$ preserve the structure of $M$. First, we choose some $x_0\in U_x$ and $y_0\in U_y$ with $\bar x_0=\bar y_0\in\pi(U_x)\cap\pi(U_y)$, hence $$x_0=g_0y_0=(\res{\pi}{U_x})^{-1}(\res{\pi}{U_y})y_0\text{ for some }g_0$$ Since $\pi(U_x)\cap\pi(U_y)$ is connected so are $(\res{\pi}{U_y})^{-1}(\pi(U_x)\cap\pi(U_y))$ and $(\res{\pi}{U_x})^{-1}(\pi(U_x)\cap\pi(U_y))$, we claim that $$g_0y=(\res{\pi}{U_x})^{-1}(\res{\pi}{U_y})y\text{ for all }y\in(\res{\pi}{U_y})^{-1}(\pi(U_x)\cap\pi(U_y))$$ Let the path $\gamma:[0,1]\to(\res{\pi}{U_y})^{-1}(\pi(U_x)\cap\pi(U_y))$ have $y_0$ and $y$ as its initial and terminal points respectively. Let $$S=\{T\in[0,1]:g_0\gamma(t)=(\res{\pi}{U_x})^{-1}(\res{\pi}{U_y})\gamma(t)\text{ for all }t\in[0,T]\}$$ Obviously $0\in S$. Then let $$T_0=\sup S$$ By the continuity of $g_0,(\res{\pi}{U_x})^{-1}$ and $\gamma$ we have $T_0\in S$. We claim $T_0=1$. If not, suppose $T_0<1$ and let $y_0'=\gamma(T_0)$, then there is a sequence $y_k=\gamma(T_0+\vare_k)$ such that $$y_k\to y_0'$$ $$(\res{\pi}{U_x})^{-1}(\res{\pi}{U_y})y_k=g_ky_k\neq g_0y_k\text{ with }g_k\neq g_0$$ By the continuity of $(\res{\pi}{U_x})^{-1}(\res{\pi}{U_y})$ we have $$g_ky_k=(\res{\pi}{U_x})^{-1}(\res{\pi}{U_y})y_k\to(\res{\pi}{U_x})^{-1}(\res{\pi}{U_y})y_0'=g_0y_0'$$ that is, $$g_kg_0^{-1}(g_0y_k)\to g_0y_0'$$ On the other hand, the continuity of $g_0$ also gives $$g_0y_k\to g_0y_0'$$ Since $G$ acts properly discontinuously without fixed points, $g_0y_0'$ has a neighborhood $U$ such that $gU\cap U=\varnothing$ for all $g\neq\id$. Hence we have $$g_0y_k\to g_0y_0'\\ \implies g_0y_k\in U\text{ for all sufficiently large }k\\ \implies g_kg_0^{-1}(g_0y_k)\notin U\text{ for all sufficiently large }k\text{ because }g_k\neq g_0$$ contradicting $g_kg_0^{-1}(g_0y_k)\to g_0y_0'$. This means $T_0=1$ and by the definition of $T_0,\gamma$ and $S$ we obtain $$g_0=(\res{\pi}{U_x})^{-1}(\res{\pi}{U_y})\text{ in }(\res{\pi}{U_y})^{-1}(\pi(U_x)\cap\pi(U_y))$$ It follows that the transition maps have the form $$\varphi_xg\varphi_y^{-1}$$ with $g$ being an automorphism that preserves the structure of $M$. It follows that $M/G$ admits an atlas and hence a structure of the same type as $M$.
The type of group action in the updated proposition is also known as a "covering space action". (A term, I think, coined by Hatcher.)
TFAE for an isometric group action of a locally compact group $G$ on a locally compact Hausdorff metric space $M$ (e.g. any manifold):
- $G$ acts properly discontinuously and freely (=without fixed points);
- $G$ acts totally discontinuously (every $x$ has a nbh $U$ with $gU \cap U \neq \varnothing \implies g=e$);
- The map $M \to M/G$ is a covering map. (And hence $M/G$ inherits the structure from $M$ if $G$ preserves that structure.)
- $G$ acts freely with discrete orbits;
and they imply that $G$ is discrete. Metrizability is only needed for 4 $\implies$ 1,2,3.
See e.g. Proposition 4 in these notes on Fuchsian groups by Pete L. Clark.