Let $X$ be your space, and note that it can be realized as a quotient of $I^2$. In particular, the usual structure of the torus is a quotient map $q:I^2 \to I^2/\sim$ by taking $(x,0) \sim (x,1)$ and $(0,y) \sim (1,y)$.
There is another quotient now $p:I^2 \to I^2/\sim$ by taking $(x_1,0) \sim (x_2,0)$ and $(x_1,1) \sim (x_2,1)$. Your space is done by doing $q$ first and then $p$ (or technically the map induced by $p$.)
However, the situation is clearer if you do $p$ first and then $q$.
Pinching two sides of the square down to a point gives a structure of $2$ zero cells, and two $1$-cells joining them, and filling this with a $2$-cell. The induced bap by $q$ glues these two $1$ cells together giving a sphere, but we still have to identify the $0$-cells, so this is a sphere with two points identified.
There is an excellent illustration that a sphere with two points identified is homotopically just $S^2 \vee S^1$ here.
Now one can apply siefert-Van kampen to obtain the result.
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
I'll presume that by the "definition of the mobius band in $[0,1] \times [0,1]$" you mean that one should identify $(0,t) \sim (1,1-t)$ for each $t \in [0,1]$.
I'll also presume that by the "point of origin" you mean the point $(0,0) \in [0,1] \times [0,1]$.
Under those presumptions, what you get is the moebius band with one point removed from its boundary circle. Like the moebius band itself --- which is homotopy equivalent to $\mathbb S^1$ --- the result of removing one point from the boundary circle is still homotopy equivalent to $\mathbb S^1$. In fact, removing a point from the boundary of any manifold-with-boundary leaves the homotopy type unchanged.
On the other hand, suppose that you had asked what you get when removing the point $\left(\frac{1}{2},\frac{1}{2}\right)$, which would be the same as removing a point from the interior of the moebius band itself. Then the result would be homotopy equivalent to $\mathbb S^1 \vee \mathbb S^1$.