My second question is unambiguous and can be answered affirmatively, as Connor Malin suggested in his comment, by cohomological means using $ \mathbb{R}P^\infty $. This is done by Dmitry Vaintrob in his answer here
https://mathoverflow.net/questions/414465/mathbbrpn-bundles-over-the-circle/414493?noredirect=1#comment1062832_414493
Furthermore, Dimtry Vaintrob notes that this result answers my first question up to homotopy and the classification is as I conjectured. However Tom Goodwillie notes in his answer to the same question that the classification is wild up to diffeomorphism because of exotic smooth spheres and therefore my conjecture about $ \mathbb{R}P^n \rtimes S^1 $ is false working up to diffeomorphism.
Indeed, a big moral of the story for me with this experience has been that my question "Classify $ \mathbb{R}P^n $ bundles over $ S^1 $" is very vague since topologists use many notions of equivalence. Including classification up to homotopy, or homeomorphism, or diffeomorphism, or even classification as bundles with a certain structure group.
I'm late to the party, though I wanted to share a quick proof using differential geometry.
Let $F$ be an orientable, connected manifold.
Any fiber bundle $F\hookrightarrow E \overset{\pi}{\to} B$ admits a connection (see for example here) this is just a splitting of the tangent bundle of the total space $TE\simeq H\oplus V$ where $V = \ker d\pi$ is the vertical subbundle (given by $\pi$) and $H$ is the horizontal subbundle (what gives the connection).
Any fiber bundle admits a connection, so we pick one and compute the first Steifel-Whitney class*:
$$ w_1(TE)= w_1(H) + w_1(V) = \pi^*(w_1(TB))+ \pi^*(w_1(E))= \pi^*( w_1(TB)+ w_1(E))\in H^1(E;\mathbb Z/2) $$
where we used the isomorphism $H \overset{d\pi}{\to} TB$ implying that $w_1(H)=\pi^*(w_1(TB))$ and that the obstruction to orient the bundle $\pi^*(E)\to E$ is $w_1(V)$ (see **)
Corollary:
i) if $B$ is orientable then the total space $E$ is orientable (as manifold) iff the bundle $E\to B$ is orientable.
ii) if $B$ is not orientable, i.d. $w_1(TB)\neq 0$, then $E$ is orientable (as manifold) iff $w_1(E) = w_1(TB)$ (note that $\pi^*: H^1(B; \mathbb Z/2) \to H^1(E; \mathbb Z/2)$ is injective because $\pi_1(E)\to \pi_1(B)$ is surjective, abelianization is right exact and $Hom(\cdot, \mathbb Z/2)$ is left exact).
*To avoid confusion, I will AVOID the shorthand notation $w_1(X) \sim w_1(TX)$, so when I write $w_1(\xi)$, it is clear that $\xi$ is a bundle.
** If $E$ is vector bundle we have that $\pi^* E \simeq V$. For a general fiber bundle, notice that if $\gamma\subset E$ is a loop, and I want to check if the bundle given by $\bigcup_{t} E|_{\pi(\gamma(t))}\to \mathbb S^1$ is orientable, then (since $F$ is orientable and connected) this is the same as checking what happens at a tangent space to the fiber at one point (does the monodromy reverse the orientation?) but this is exactly the same as looking at the monodromy of $\gamma^*(V)\to \mathbb S^1$.
Best Answer
The total space $TM$ of tangent bundle to any manifold $M$ (orientable or not) is orientable, see here.
The total space $UTM$ of the unit tangent bundle is a hypersurface in $TM$ which admits a nowehere vanishing normal vector field (at each point $(x,v)\in UTM$ take the normal vector $((x,v),v)$).
Thus, $UTM$ is a cooriented hypersurface in an orientable manifold, hence, is itself orientable.