Lemma: Given the commutative diagram
$$\begin{array}{ccccccccc} \widetilde{X} & \\
\downarrow{\small{p}} & {\searrow}^{q} \\
X_1 & \!\!\!\!\! \xleftarrow{p_1} & \!\!\!\! X_2\end{array}$$
where $p_1, p$ are covering maps, then so is $q$, where $X_1, X_2, \tilde{X}$ are all path-connected and locally path-connected.
Proof:
$q$ is surjective: $\sigma$ be a path in $X_2$ from $x_0$ and $x$. Pushforward by $p_1$ to get a path $p_1 \circ \sigma$ in $X_1$ from $p_1(x_0) = x_0'$ to $p_1(x)$. Lift to $\widetilde{X}$ to get a path $\widetilde{\sigma}$ starting from some point $x_0''$ in the fiber over $x_0'$. Pushforward by $q$ to get path $q \circ \widetilde{\sigma}$ starting at $x_0$. Uniqueness of path-lifing says $q \circ \widetilde{\sigma} \simeq \sigma$, so that $q$ maps the endpoint of $\tilde{\sigma}$ to the endpoint $x$ of $\sigma$. As $X_2$ is path connected, we can apply this argument for all $x \in X_2$ to prove $q$ is surjective.
$q$ is a covering map: Pick $x \in X_2$. Pushforward by $p_1$ to get $p_1(x)$ in $X_1$. There is a path-connected neighborhood $\mathscr{U}$ of $p_1(x)$ evenly covered by $p_1$ and $p$ (take neighborhoods evenly covered by $p_1$ and $p$ and take intersection). $\mathscr{V}$ be the slice in $p_1^{-1}(\mathscr{U})$ containing $x$. $\{\mathscr{U}_\alpha\}$ be the slices in $p^{-1}(\mathscr{U})$. $q$ maps each slice $\mathscr{U}_\alpha$ to distinct slices in ${p_{1}}^{-1}(\mathscr{U})$. $q^{-1}(\mathscr{V})$ is then union of slices in $\{\mathscr{U}_\alpha\}$ which are mapped homeomorphically onto $\mathscr{V}$. I claim all $\mathscr{U}_\alpha$ are mapped homeomorphically on $\mathscr{V}$ by $q$. This can be proved slicewise, recalling that given a commutative diagram with any two arrows as homeomorphism, so is the third. $\blacksquare$
If $\widetilde{X}$ is simply connected, $p : \widetilde{X}\to X$ the universal cover, $p_1 : X_2 \to X_1$ a covering map, then as $p_*(\pi_1(\widetilde{X}))$ fits inside ${p_1}_*(\pi_1(X_2))$, being the trivial group, we can lift $p$ to $\tilde{p} : \widetilde{X} \to X_1$. By previous discussion, $\tilde{p}$ is a covering map, since it fits inside a commutative diagram like above. Thus, $\widetilde{X}$ covers $X_2$, as desired.
I think only an odd number of sheets are possible.
This is a great example where the general theory leads to interesting computational results in particular cases: we can determine possible coverings of the form $M\to M$ by first determining all connected coverings of $M$, and then detecting which ones have total space homeomorphic to $M$.
Classification Theorem: For any path-connected, locally path-connected, semi-locally simply-connected space $X$ there is a bijection between isomorphism classes of connected covering spaces of $X$ and conjugacy classes of subgroups of $\pi_1(X)$. (See for example Theorem 1.38, page 67.)
This works by constructing a universal covering $\tilde{X}\to X$ so that the quotients $\tilde{X}/H$ represent all the connected coverings of $X$ as $H$ varies over conjugacy classes. The covering $\tilde{X}$ is characterized up to covering space isomorphism by being simply-connected.
Universal covering space of $M$: Recall $M\sim S^1$ so $\pi_1(M)\cong \mathbb{Z}$. The universal covering space of $M$ is $\mathbb{R}\times [0,1]$ and the action of $\mathbb{Z}$ is given by $n\cdot (x, t)= \big(x+n, f^{n}(t)\big)$ for $n\in\mathbb{Z}$ and where $f\colon [0,1] \to [0,1]$ is the "flip" homeomorphism given by $f(t)= 1-t$. (Visually, think about $\mathbb{R}\times[0,1]$ as an infinite strip of tape that you're applying to the Möbius strip, which is alternating "front" and "back" sides.)
Quotients of $\tilde{M}$: Every connected covering of $M$ is a quotient of the form $(\mathbb{R}\times [0,1])/n\mathbb{Z}$. For each $n$ a fundamental domain of the quotient is $[0,n]\times[0,1]$, and the quotient only depends on how we identify the subspaces $\{0\} \times [0,1]$ and $\{n\}\times [0,1]$. When $n$ is odd then $f^{n} = f$ so we identify the ends using a flip, and hence the quotient is homeomorphic to $M$; on the other hand if $n$ is even then $f^{n}=id$ and so the quotient is actually the cylinder $(\mathbb{R}/n\mathbb{Z})\times [0,1]$.
Since these quotients make up all of the possible connected coverings of $M$, it follows that coverings of the form $M\to M$ can have any odd number of sheets.
Best Answer
$X$ is a connected space.
Let $|A|$ denote the cardinality of set $A$.
Let $U$ be an evenly covered open set (of $X$) and let open sets(of $\tilde{X}$ ) ${U_α}$, $α \in J$, an index set, be the partitions of $p^{-1}(U)$. Also $p$ restricted $U_α$ is a homeomorphism between $U_α$ and $U$.
Let $x \in U$. Note that $p^{-1}$ {$ x $} $\bigcap U_α$ has exactly one element. Hence $|p^{-1}$ {$ x $}$|$ = $|J|$. This means that for $a,b \in U$, $|p^{-1}$ {$a$}$|$ = $|p^{-1}$ {$b $}$|$. Now let $a \in X$.
$A$={$x \in X$; $|$$p^{-1}$ {$x$}$|$ =$|$$p^{-1}$ {$a$}$|$ }. By above remarks both $A$ and $X\backslash A$ are open. Since $X$ is connected this means $X\backslash A$ is empty. Hence $|p^{-1}$ {$ x $}$|$ is same for every $x \in X$.