Hint for the first question. Given the cone $\Gamma X$ of a topological space $X$, you can cut it in the middle : $\Gamma X \simeq X_1 \cup X_2$ with $X_1 \simeq \Gamma X$, $X_2 \simeq X\times[0,1]$ and $X_1 \cap X_2 \simeq X$.
Given $f,g \colon X \to Y$ homotopic by $H \colon X \times [0,1] \to Y$, you could now be able to define a map $$M_g \to M_f$$ which will be (easily) a homotopy equivalence.
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
Here is a result with a bit of point set.
Point set fact: if $X,Y$ are hausdorff and homeomorphic, then they have homeomorphic one point compactifications.
With this, you can show that the one point compactification of the mobius band is hoeomorphic to $\mathbb RP^2$. This can be seen explicitly by considering the punctured projective plane.
The one point compactification of the cyllinder is a sphere with two points identified.
You can either stop here if you are satisfied with orientability (or embeddings into $\mathbb R^3$.)
Otherwise, note that the latter space is homotopic to $S^2 \vee S^1$.
Now, you have the job of distinguishing $S^2 \vee S^1$ and $\mathbb RP^2$ up to homotopy is not so bad. For example, what are the possible maps $\pi_1(\mathbb RP^2) \to \pi_1(S^2 \vee S^1)$? Or said differently, there is a loop of order $2$ in $\mathbb RP^2$, is there anything like that in $S^2 \vee S^1$? So on, and so forth.