$\mathbf P^1(\mathbb R)$ is a closed curve, so its tangent bundle is trivial. There is not much to be imagined, really, in that case! In general, what kind of answer do you expect for your «what is the tangent bundle of $\mathbf P^n$?». It is the tangent bundle of $\mathbf P^n$...
Regarding your last question: if a group $G$ acts properly discontinuously and differentiably on a manifold $M$, then $M/G$ is a manifold, $G$ acts on $TM$ in a natural way (also properly discontinuously, and morover linearly on fibers), and $(TM)/G$ is "the same thing" as $T(M/G)$.
First understand what we mean by an orientation for $\mathbb{R}^3$. This is tied to the idea of a right-handed set of vectors: we say $\{v_1,v_2, v_3 \}$ is a right-handed set of vectors if $\text{det}(v_1|v_2|v_3)>0$. Geometrically, this means that $v_3$ is on the same side of the plane spanned by $v_1$ and $v_2$ as $v_1 \times v_2$. We are most familar with the case of an orthonormal right-handed frame where we simply insist $v_1 \times v_2 =v_3$. In two dimensions, the condition $\{ v_1,v_2 \}$ be right-handed again is captured by $\text{det}(v_1|v_2)>0$. Geometrically, the condition in two-dimensions means that the second vector is obtained from the first by a Counter-Clockwise rotation. In all cases, an orientation allows us to decide which way is up once all the other ways are fixed. To be slighly more precise, if we fix $n-1$ coordinates in $\mathbb{R}^n$ then an orientation will provide us a framework in which we can decide what the positive direction is for the remaining $n$-th coordinate. This is equivalent to insisting there be an ordered set of vectors $\{v_1,v_2,\dots v_n \}$ for which $\text{det}[v_1|v_2|\cdots|v_{n-1}|v_n]>0$.
Details of the choice: ok, suppose $v_1,\dots , v_{n-1}$ are fixed and you are given $a,b$ vectors in $\mathbb{R}^n$ then $\text{det}[v_1|v_2|\cdots|v_{n-1}|a]>0$ whereas $\text{det}[v_1|v_2|\cdots|v_{n-1}|b]<0$. Then, in terms of the given orientation $\{v_1,v_2,\dots v_n \}$ we see $a$ points in the upward-direction whereas $b$-points in the downward-direction. Of course, this comment is relative to the coordinate system defined by the span of the orientation.
Relation to wedge product:
$$ v_1 \wedge v_2 \wedge \cdots \wedge v_n = \text{det}[v_1|v_2|\cdots|v_{n}] e_1 \wedge e_2 \wedge \cdots \wedge e_n. $$
Where $(e_i)_j = \delta_{ij}$ defines the standard basis. The coefficient of the $n$-form will be positive iff the set of vectors $\{v_1,v_2, \dots, v_{n}\}$ shares the same orientation as the standard basis. Moreover, all such related bases form the so-called standard orientation of $\mathbb{R}^n$.
Now, for the case of a manifold it is much the same, however, we have to consider the derivations which fill $T_pM$ at various $p$. The natural orientation given by a coordinate chart $(x^i)$ at $p \in M$ is simply the ordered $n$-tuple (using $\partial_i = \frac{\partial}{\partial x^i}$)
$$ \{ \partial_1|_p, \partial_2|_p, \dots, \partial_n|_p \}$$
dual to these $dx^1, dx^2, \dots dx^n$ form the volume form:
$$ vol=dx^1 \wedge dx^2 \wedge \cdots \wedge dx^n$$
Given this, we can judge if $v_1,v_2, \dots v_n$ is a coherent orientation to the one which is naturally induced from the coordinate derivations. Simply check:
$$ vol(v_1,v_2, \dots , v_n) > 0 ? $$
Now, if there is another coordinate system $(y^j)$ also defined at $p$ then we can ask if
$$ \{ \partial/\partial y^1,\partial/\partial y^2, \dots, \partial/\partial y^n \}$$
forms a coherent orientation with the one we already induced from $(x^i)$. The chain rule connects the $x$ and $y$ coordinate derivations and when we feed that to the volume form we find the determinant of the transition function appears. Therefore, two coordinate systems provide coherent orientations if their transition functions have Jacobians with positive determinant.
All of this said, it seems that the phrase you quote:
If we can assign an orientation to each point on a manifold M in such a way that the orientations as any two sufficiently near points on M are coherent, . .
could apply to what I describe above. But, the larger idea concerns a curve so almost certainly the discussion above about coordinate-induced orientations is not the real answer to the question.
What about a curve? Morita talks about choosing an orientation along the curve. I'm not entirely sure which construction he has in mind. But, here's a possibility: given a curve we can frame the curve using the tangent vector, and the change in the tangent vector etc.. well, in the abstract, how to frame a curve? I leave to your imagination, but, in many cases it can be done. So, there is a natural way to assign $n$-vectors $f_1(t), \dots , f_n(t)$ to a given curve at point $\gamma(t)$. Suppose the curve is closed such that after time $T$ the curve returns to the initial point $p=\gamma(0)=\gamma(T)$. Then, we can compare the orientations of $f_1(0), \dots , f_n(0)$ and $f_1(T), \dots , f_n(T)$ and see if they are compatible (coherent). In particular, feed both to the volume form at $p$ and see that they share the same signed result. If you did this for the tangent, normal frame field for a curve on the Mobius band then you'd find the orientation carried by the frame of the curve was not coherent when you go around the band once.
Best Answer
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.