Let $M\subset \Bbb R^3$ be a $2$ dimensional submanifold with boundary (say the Möbius strip) and let $p\in M$ be a boundary point. Let $U$ be an open ball of $\Bbb R^3$ centered on $p$, like in your definition of embedded submanifold. Then $U\cap M$ will be an open neighbourhood of $p$ in $M$, diffeomorphic to an open half-disk $D$ in $\Bbb R^2$ ("diameter" included). Finally, observe that there is no open subset of $\Bbb R^2$ diffeomorphic to $D$, so there is no $V\subseteq \Bbb R^3$ open such that $\phi(U\cap M)=V\cap \Bbb R^2$.
Note that the difference between a manifold with an one without boundary lies exactly in the fact that every point of the former has a neighbourhood diffeomorphic to $\Bbb R^n$, while this fails for some points of the latter, which admit open neighbourhoods diffeomorphic to an open half-ball in $\Bbb R^n$.
They are homeomorphic, and they are both orientable.
It’s not clear whether you’re considering just the surfaces of these objects, or the whole solid objects, but in either case, the answer is the same. I’ll assume in the following that you mean the surfaces, but it adapts straightforwardly to the solid bodies as well. I’m leaving a lot of details out here; if you would like me to elaborate on anything, let me know.
To get the homeomorphism, first pick nice parametrisations of the torus and the fattened Möbius band. $\newcommand{\x}{\mathbf{x}}\newcommand{\y}{\mathbf{y}}\newcommand{\R}{\mathbb{R}}$The torus can be parametrised by pairs $(\theta,\x)$, where $\theta \in [0,2 \pi)$, and $\x \in \R^2$, $|\x| = 1$, i.e. $x$ is a point on the standard unit circle. Here $\theta$ represents the “major” angle, i.e. the angle on the large-diameter circle round the origin, and $\x$ represents the point on the small-diameter circle.
The fattened band is exactly the same idea, but a little tricker to write down. Let $Q_\theta$, for $\theta \in [0,2\pi]$, be the unit square in $\R^2$ rotated by $\theta/4$ radians. So as $\theta$ varies from $0$ to $2\pi$, $Q_\theta$ rotates by a quarter-turn, and ends up back where it started. Now the fattened band can be parametrised by pairs $(\theta,y)$, where $y \in Q_\theta$.
Now one direction of the homemorphism, from the band to the torus, is described easily by sending $(\theta,\y)$ to $(\theta, \frac{\y}{|\y|})$.
Once one has the homeomorphism, it follows that since the torus is orientable, so is the fattened Möbius band.
One take-home here is the Möbius band is very different from the fattened Möbius band; the non-orientability of the former doesn’t imply anything about the latter.
Best Answer
In your non-orientable surface there must be a simple, closed loop that "switches orientation" (i.e. if you go along the loop, using charts and positive Jacobian to keep your orientation all the way around, then you will have switched orientation by the time you go one full lap). A thin neighborhood of that loop will be a Möbius band.
As for why such a loop exists, take a finite cover of charts of your surface (which exists because of compactness). Now, start with one of them $U_1$. Next, pick a second chart $U_2$ that intersects $U_1$. If they have negative Jacobian on the overlap, invert the orientation of $U_2$. You have now picked an orientation on $U_1\cup U_2$. Continue with $U_3$, which intersects $U_1\cup U_2$, to get an orientation on $U_1\cup U_2\cup U_3$. And so on.
At some step $n$, this will necessarily fail, because the surface is non-orientable and our cover is finite. In practice, what will happen is that the overlap $\left(\bigcup_{i<n}U_i\right)\cap U_n$ has several connected components, and among these there are two components with opposite Jacobian sign. No matter what orientation you choose for $U_n$, one of those two overlaps will be orientation reversing.
Now take a point $p$ in one of those two connected components, and a point $q$ in the other. Connect them by a path through $U_n$ and by a path through $\bigcup_{i<n}U_i$. Together these two paths make a loop that switches orientation.