One way of thinking about orientability of an arbitrary rank $k$ vector bundle $E$ is to think about it analogously to the orientability of the tangent bundle.
Recall that the definition of an orientable tangent bundle is that the manifold is orientable, which is equivalent to the existence of a nowhere vanishing top form, aka a volume form. We can do something similar to the vector bundle $E$. If $E$ has rank $k$, we can form the top exterior power $\wedge^kE$. Then we say, analogous to the case of the tangent bundle, that the vector bundle $E$ is orientable iff there exists a nowhere vanishing section of $\wedge^kE$.
There is no direct relation between the orientability of a vector bundle and the base manifold. The reason that there is one between the tangent bundle is that the two notions (orientable manifold and orientable tangent bundle) mean exactly the same thing! So to ask for the relation of orientability of an arbitrary vector bundle and of the base manifold is to ask for the relation between an arbitrary vector bundle and the tangent bundle; and in general there isn't one.
(Just to illustrate: consider the Moebius strip formed by taking $[0,1]\times[0,1]$ and giving a half-twist. If you embed the Moebius strip in $\mathbb{R}^3$, and pull-back the tangent bundle of $\mathbb{R}^3$ to the strip, it is an orientable vector bundle when the Moebius strip itself is not orientable. On the other hand, consider the image of $S^1$ in the midline of the Moebius strip. Its normal bundle relative to the Moebius strip is not orientable, while $S^1$ is.)
About that double tangent bundle: if you treat $TTM$ as a bundle over $TM$, then for arbitrary $M$, since $TM$ is orientable as a manifold, clearly $TTM$, as its tangent bundle, is orientable as a bundle.
Let me add one more thing: if you are familiar with the proof that $TM$ is always orientable as a manifold, you can use the same method to prove the following theorem:
Theorem Let $E$ be a vector bundle over $M$. Consider the statements (i) $M$ is orientable as a manifold (ii) $E$ is orientable as a manifold (iii) $E$ is orientable as a vector bundle. Any two of the statements being true will imply the third.
If $M$ is a manifold and $\hat M$ its universal cover, $\hat M$ is endowed with a differentiable structure such that $p:\hat M\rightarrow M$ is differentiable. For every $x\in M$, there exists a neighborhood $U$ of $x$ such that $U$ is a domain chart $\phi:U\rightarrow\mathbb{R}^n$ and $p^{-1}(U)=\bigcup V_i$ and $p_{\mid V_i}\rightarrow U$ is an homeomorphism, the differentiable structure is defined by supposing that $p_i$ is a diffeomorphism which induces the chart $\phi\circ p_i$ on $\hat V_i$.
Now look at the action of an element $\gamma\in\pi_1(M)$ on $\hat M$, i.e. a covering transformation. We will prove that this action is already smooth. Let $p(y)=x$. There exist $i$ and $j$ such that $y\in V_i$ and $\gamma . y\in V_j$,
However, the map $(\phi\circ p_j) \circ \gamma \circ (\phi\circ p_i)^{-1}$ is the identity, this implies that action of $\gamma$ is differentiable.
This proves that every covering transformation is smooth.
Best Answer
If $\pi : N \to M$ be a smooth covering map. Then $\pi$ is a local diffeomorphism so it follows that $\pi^*TM \cong TN$. The characteristic classes of $N$ are therefore the pullbacks of the corresponding classes of $M$ by $\pi$. Another corollary is that if $TM$ is trivial, then so is $TN$, but the converse is not true as you've observed. Even if $N = M'$ and the covering is finite, $TM$ may not be trivial, e.g. $S^2\times S^5$ has trivial tangent bundle, but $S^2\times\mathbb{RP}^5$ does not since $w_2(S^2\times\mathbb{RP}^5) \neq 0$.
Regarding triviality of $T\mathbb{RP}^n$, from the above we know that if $T\mathbb{RP}^n$ is trivial, then so is $TS^n$ and hence $n \in \{1, 3, 7\}$. However, it is not clear if all three values of $n$ yield a trivial bundle. This turns out to be true. One way to see this is to give a basis of vector fields $\{V_i\}_{i=1}^n$ on $S^n$ for which $V_i(-x) = -V_i(x)$, so they descend to a basis of vector fields on $\mathbb{RP}^n$; see this answer for details.