Manifold Orientability Definition

Let me speak of a vector bundle $E \to M$ over a smooth manifold $M$, and let $F$ be the typical fiber of the bundle $E$. In particular, for each point $p \in M$ we can find an open neighborhood $U \ni p$, such that $E|_U$ is diffeomorphic to $U \times F$. A choice $\varphi \colon E|_U \to U \times F$ is called a local trivialization of $E$.

The points $p,q \in M$ are "sufficiently near" if there is an open set $U \subseteq M$, such that $p,q \in U$ and there is a local trivialization $\varphi \colon E|_U \to U \times F$.

In the trivial vector bundle $V \times F$ all the fibers are canonically isomorphic via the map $\tau \colon (p,v) \mapsto (q,v)$. Using this we can compare the orientations of the fibers $E_p$ and $E_q$ for "sufficiently near" points. If the canonical isomorphism $\tau$ is orientation-preserving for these points, the chosen orientations in the fibers are called coherent.

One can find some insights how to make the full statements, for instance, in this discussion, and in textbooks, e.g. Loring Tu, "An Introduction to Manifolds", p. 240.

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.