I was wondering if it is sufficient for a compact (i.e. Hausdorff) smooth manifold $M$ to have a free group action of a finite group $G$ in order to conclude that $M/G$ is a compact smooth manifold?
This is definitely wrong if $M$ is not compact, but I suspect that it might be true for such manifolds.
I think that for finite $G$, $M/G$ is a smooth manifold, regardless of any compactness assumptions. For a general discrete group $G$ acting on a manifold $M$, you need the action to be free and properly discontinuous in order to make $M/G$ into a smooth manifold (and $M\to M/G$ into a covering). The second condition means that for each $x\in M$ there is a neighbourhood $U$ of $x$ in $M$ such that $gU\cap U\neq\emptyset$ is only possible for $g=e$. This means that the projection $M\to M/G$ restricts to a homoemorphism on $U$ and you can use this to construct charts on $M/G$. For a finite group $G$ this condition is always satisfied, an appropriate neighbourhood is easily constructed using that $M$ is Hausdorff.