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 answer is "yes", and the result can be found in Bourbaki's General Topology, Part I, Exercise 19 to $\S$ 10 of Chapter 1,
on page 153 of the English translation published by Springer.
The exercise has a hint, but it took me quite a time to fully elaborate the proof.
So I try to present the argument with some more details included.
Let us first make clear some terminology.
A neighbourhood of a set $A$ is any set containing an open set containing $A$.
A space $X$ is compact if any open cover of $X$ has a finite subcover.
A space is locally compact if it is Hausdorff and any point has a compact neighbourhood;
it is $\sigma$-compact if it is a countable union of compact sets.
A $\sigma$-locally compact space is a space which is both $\sigma$-compact and locally compact.
A normal space is a space in which any two disjoint closed sets have disjoint neighbourhoods.
It is well known that every compact Hausdorff space is normal, and
in a normal space, any two disjoint closed sets have disjoint closed neighbourhoods.
A normal space in which every singleton is closed, is necessarily Hausdorff.
If $X$ is compact and $Y$ is arbitrary then the projection to the second coordinate $\textit{pr}_2\colon X\times Y\to Y$ is a closed mapping.
It is also known that every $\sigma$-locally compact Hausdorff space $X$ can be expressed as $\bigcup_{n\in\omega}X_n$,
where $X_n$ are open, $\overline{X_n}$ are compact, and $\overline{X_n}\subseteq X_{n+1}$, for every $n$.
Let $E$ be an equivalence relation on $X$.
Denote by $[A]$ the saturation of a set $A\subseteq X$, that is, $[A]=\{y\in X\!:(\exists x\in A)\,(x,y)\in E\}$.
If $Y\subseteq X$ then $E_Y=E\cap(Y\times Y)$ is an equivalence relation on $Y$.
The saturation of a set $A\subseteq Y$ with respect to $E_Y$ is $[A]\cap Y$.
If $A,B\subseteq Y$ are disjoint sets saturated with respect to $E_Y$ then their saturations $[A]$, $[B]$ with respect to $E$ are disjoint as well.
Lemma.
Let $X$ be a compact Hausdorff space and let $E$ be an equivalence relation on $X$ such that $E$ is a closed subset of $X\times X$.
Then the quotient mapping $q\colon x\mapsto[x]$ is closed.
Proof.
Let $A$ be a closed subset of $X$.
Then $A$ is compact and $[A]=\textit{pr}_2(E\cap(A\times X))$.
Since $E\cap(A\times X)$ is closed in $A\times X$ and $\textit{pr}_2$ is a closed mapping, $[A]$ is a closed subset of $X$.
Hence $q$ is closed.
q.e.d.
Lemma.
Let $X$ be a normal space and let $E$ be an equivalence relation on $X$ such that the quotient mapping $q\colon x\mapsto [x]$ is closed.
Then the quotient space $X/E$ is normal.
Proof.
Let $A,B$ are disjoint closed saturated subsets of $X$.
Since $X$ is normal, there exist disjoint open sets $U,V$ such that $A\subseteq U$ and $B\subseteq V$.
Then $[X\setminus U]$ and $[X\setminus V]$ are closed saturated sets disjoint from $A$ and $B$, respectively,
hence $U'=X\setminus [X\setminus U]$ and $V'=X\setminus [X\setminus V]$ are open saturated neighbourhoods of $A$ and $B$,
respectively.
If $z\in U'\cap V'$ then $[z]$ is disjoint from both $X\setminus U$ and $X\setminus V$, hence $z\in U\cap V$, a contradiction.
It follows that $U',V'$ are disjoint,
so any two disjoint closed subsets of $X/R$ have disjoint open neighbourhoods.
q.e.d.
Theorem.
Let $X$ be a $\sigma$-locally compact Hausdorff space and let $E$ be an equivalence relation on $X$ such that $E$ is a closed
subset of $X\times X$. Then the quotient space $X/E$ is normal and Hausdorff.
Proof.
Let $X=\bigcup_{n\in\omega}X_n$, where $X_n$ are open sets such that $\overline{X_n}$ are compact and
$\overline{X_n}\subseteq X_{n+1}$ for every $n$.
For every $n$ let us consider the quotient space $Y_n=\overline{X_n}/E_n$, where $E_n=E\cap(\overline{X_n}\times\overline{X_n})$.
Since $\overline{X_n}$ is a compact Hausdorff space, the corresponding quotient mapping $q_n\colon\overline{X_n}\to Y_n$ is closed and hence the space $Y_n$ is normal.
Let $A,B$ be disjoint closed saturated subsets of $X$.
We prove that there exist disjoint open saturated sets $U,V$ such that $A\subseteq U$ and $B\subseteq V$.
For $n\in\omega$, define sets $A_n$, $B_n$, $U_n$, $V_n$ by induction as follows.
Start with $A_0=A\cap\overline{X_0}$, $B_0=B\cap\overline{X_0}$.
In $n$-th step, assume that $A_n$, $B_n$ are disjoint closed subsets of $\overline{X_n}$ saturated with respect to $E_n$.
Assume also that $A\cap\overline{X_n}\subseteq A_n$ and $B\cap\overline{X_n}\subseteq B_n$.
Then, by normality of $Y_n$, there exist disjoint relatively open subsets $U_n$, $V_n$ of $\overline{X_n}$ such that $A_n\subseteq U_n$, $B_n\subseteq V_n$, $\overline{U_n}\cap\overline{V_n}=\emptyset$, and $U_n$, $V_n$, $\overline{U_n}$, $\overline{V_n}$ are saturated with respect to $E_n$.
We have $\overline{V_n}\subseteq\overline{X_n}$, hence $$A\cap\overline{V_n}\subseteq A\cap\overline{X_n}\cap\overline{V_n}\subseteq A_n\cap\overline{V_n}\subseteq U_n\cap\overline{V_n}=\emptyset,$$ similarly $B\cap\overline{U_n}=\emptyset$.
Since $A,B$ are saturated with respect to $E$, we also have $A\cap\big[\,\overline{V_n}\,\big]=B\cap\big[\,\overline{U_n}\,\big]=\emptyset$.
Let us take $A_{n+1}=\big(A\cup\big[\,\overline{U_n}\,\big]\big)\cap\overline{X_{n+1}}$, $B_{n+1}=\big(B\cup\big[\,\overline{V_n}\,\big]\big)\cap\overline{X_{n+1}}$.
It is easy to check that $(A\cap\overline{X_{n+1}})\cup U_n\subseteq A_{n+1}$, $(B\cap\overline{X_{n+1}})\cup V_n\subseteq B_{n+1}$, and that $A_{n+1}$, $B_{n+1}$ are closed and saturated with respect to $E_n$.
Since $\overline{U_n}$, $\overline{V_n}$ are disjoint subsets of $\overline{X_n}$ saturated with respect to $E_n$, their saturations $\big[\,\overline{U_n}\,\big]$, $\big[\,\overline{V_n}\,\big]$ with respect to $E$ are disjoint as well, and we obtain that $A_{n+1}\cap B_{n+1}=\emptyset$.
Hence we can proceed with the induction.
This way we can define increasing sequences $\{U_n\}_{n\in\omega}$, $\{V_n\}_{n\in\omega}$ such that $U_n$, $V_n$ are relatively open subsets of $\overline{X_n}$, $A\cap\overline{X_n}\subseteq U_n$, $B\cap\overline{X_n}\subseteq V_n$, $\overline{U_n}\cap\overline{V_n}=\emptyset$, and $U_n$, $V_n$, $\overline{U_n}$, $\overline{V_n}$ are saturated with respect to $E_n$.
Set $U=\bigcup_{n\in\omega}U_n$, $V=\bigcup_{n\in\omega}V_n$.
If $x\in U$ then there exists $n$ such that $x\in U_n\cap X_n\subseteq U$, where $U_n\cap X_n$ is open, hence $U$ is open.
Also, if $x\in[U]$ then there exists $n$ such that $x\in\big[\,\overline{U_n}\,\big]\cap\overline{X_{n+1}}$, hence $x\in U$.
So $U$ is an open set saturated with respect to $E$, similarly for $V$.
Finally, $U$ and $V$ are disjoint since if $x\in U\cap V$ then there exists $n$ such that $x\in U_n\cap V_n$, which is impossible.
We have proved that the quotient space $X/E$ is normal.
To see that it is also Hausdorff it suffices to show that every singleton in $X/E$ is a closed set.
This is clear since if $E$ is a closed subset of $X\times X$ then every equivalence class is closed.
q.e.d.
Best Answer
The idea is to use separating neighbourhoods for representatives of classes $[x] \neq [y]$ in $M$. We can split up the proof in three case: a) $x,y$ both lie in the interior of $X$; b) One of $x$,$y$ lies in the interior of $X$ and the other lies on the boundary; c) Both $x$ and $y$ lie on the boundary of $X$.
a) In this case, pick separating neighbourhoods $U,V$ for $x,y$ respectively that are small enough that they don't touch the boundary. Then since the equivalence relation leaves the interior of $X$ untouched, $\pi(U) \cap \pi(V) = \pi(U \cap V) = \emptyset$.
b) Without loss of generality, we may assume that $x \in \partial X$ and $y \not\in \partial X$. We may find an $\epsilon > 0$ small enough so that $y$ is not contained in the fattened boundary $U = [0,\epsilon]\times \mathbb R \cup [1-\epsilon,1] \times \mathbb R$. Since the complement of $U$ is open in $X$, we may then find a neighborhood $V$ of $y$ that also doesn't intersect $U$. The equivalence relation will leave $V$ intact and perturb the part of $U$ that lies on the boundary, and overall we will have $$\begin{align}\pi(U) \cap \pi(V) &= (\pi(\partial X) \cup \pi(U \setminus \partial X)) \cap \pi(V) \\&= (\pi(\partial X) \cap \pi(V)) \cup (\pi(U \setminus \partial X) \cap \pi(V))\\ &= \emptyset \cup \pi((U \setminus \partial X) \cap V)\\&= \emptyset.\end{align} $$
c) If $x,y \in \partial X$, we have to make sure that the separating neighbourhoods in $X$ are small enough so that they don't get squished together in $M$. For example, a tall and skinny open neighbourhood of $(0,1)$ that contains $(0,-1)$ will, down in $M$, intersect a tall and skinny open neighbourhood of $(1,1)$ that contains $(1,-1)$, even though they don't intersect in $X$. The idea, then, is to make sure that the neighbourhoods are not too tall. To this end, choose the representatives $x,y$ so that their first coordinate is $0$, i.e., they both lie on the left boundary of $X$. Then any $U,V$ that separate $x,y$ and don't intersect the right boundary of $X$ will do the job, because the parts of $U,V$ that don't intersect the left boundary boundary will be unchanged by the projection, and the parts that intersect the left boundary won't get mixed up by the projection (as $\pi$ is injective when restricted to $X$ minus the right boundary). I'll leave it to you to write this down as formally as you want to.