As I'm sure you know, the category of smooth (topological?) manifolds is one of those categories where the objects are very nice but the category itself is terrible. I cannot describe the number of times I've heard the algebraic geometers curse the smooth category.
I am not certain this is a total classification, but From Lee's Introduction to Smooth Manifolds, Theorem 9.19:
If $\tilde M$ is a connected smooth manifold and $\Gamma$ is a discrete group acting smoothly, freely, and properly on $\tilde M$, then the quotient $\tilde M/\Gamma$ is a topological manifold and has a unique smooth structure such that $\pi: \tilde M \to \tilde M/\Gamma$ is a smooth covering map.
The manifold portion of this comes from the Quotient Manifold Theorem:
If $G$ is a Lie group acting smoothly, freely, and properly on a smooth manifold $M$, then the quotient space $M/G$ is a topological manifold with a unique smooth structure such that the quotient map $M \to M/G$ is a smooth submersion.
And then applying this to the (zero-dimensional) Lie group of deck transformations.
Edit: The proof of the Hausdorff property is very similar to @useruser43208's response, and uses the properness of the action. Take the orbit set
$$ \mathcal O = \{ (g\cdot p,p): g \in G, p \in M \} \subseteq M \times M$$
which is closed under the properness assumption. Any two distinct points $\pi(p)$ and $\pi(q)$ in the image of the quotient map $\pi: M \to M/G$ must have arisen from distinct orbits, so $(p,q) \notin \mathcal O$. Hence we may find a product neighbourhood $U_p\times U_q \subseteq M \times M$ of $(p,q)$ disjoint from $\mathcal O$, hence $\pi(U_p)$ and $\pi(U_q)$ are separating open neighbourhoods (since $\pi$ is open).
The manifold $\Rightarrow \sigma-$compact looks fine. The other direction probably also works, but you can prove it much more directly/easier.
It suffices to show $X$ the space is second countable. Let $K_n$ be the covering of $X$ by compact sets, and assume they're all nonempty. For each $x\in K_n$, there is an open neighborhood $U_{x,n}$ which is homeomorphic to an open ball in $\mathbb{R}^m$. Then we have $K_n\subset \bigcup_{x\in K_n} U_{x,n}$. By compactness, there are finitely many $U_{x,n}$ which cover each $K_n$. By unioning these all together for each $n$ gives us a countable covering for $X$ by open sets which are homeomorphic to open balls in $\mathbb{R}^m$. Each of these sets has a countable basis. The countable union of these sets is also a countable set, so it remains to show that this set is a basis, which is easy.
Best Answer
Tu allows manifolds having connected components of different dimensions. He explicitly says it in this post. Usually people talk about a space being "locally $\Bbb R^n$" or "locally Euclidean of dimension $n$" as opposed to just "locally Euclidean", as he does. But it is not hard to show that for each $n \geq 0$, the set $$\{ x \in M \mid x \mbox{ has an open neighborhood homeomorphic to }\Bbb R^n \}$$is both open and closed in $M$. So this means that the dimension is well defined on each connected component of $M$.