I'm reading Warner. "Foundations of Differentiable Manifolds and Lie Groups." p. 138. I don't get the statement in the definition of orientable manifolds.
4.1 Definitions $\;$ (the preface omitted) Let $M$ be a connected differentiable manifold of dimension $n$. We chall call $M$ orientable if it is possible to choose in a consistent way an orientation on $M_m^*$ for each $m\in M$. More precisely, let $O$ be the "$0$-section" of the exterior $n$-bundle $\Lambda_n^*(M)$; that is, $$O = \cup_{m\in M} \{0\in\Lambda_n(M_m^*)\}.$$ Then since each $\Lambda_n(M_m^*)-\{0\}$ has exactly two components, it follows easily that $\Lambda_n^*(M)-O$ has at most two components. We say that $M$ is orientable if $\Lambda_n^*(M)-O$ has two components; and if $M$ is orientable, an orientation on $M$ is a choice of one of the two components of $\Lambda_n^*(M)-O$. (the rest omitted)
Why does $\Lambda_n^*(M)-O$ have at most two components?
Let there be 3 points p,q and r in the fibers above s,t and u which are points of M. M is connected so you can give paths between all 3.
There is a path from s to t. This lifts to a path from p to something projecting to t. Call this w. You can build this path by using the local triviality over small neighborhoods $U \subset M$. If we get something in the same component of $\Lambda_n (M_t^*)$. If this is in the same component as q then we could get a path from p to q.
So if we want p,q and r to all be in different components, we can assume that we landed in the wrong component.
Repeat this for t to u. But this time start from w instead of q. If we can connect from w to r, then we get a path from p to r. If we are in the wrong component for r, then do a fiberwise reflection. Now it is a path from $-w$ to the correct component for r. $-w$ is in the same component as q so this gives a path from q to r.
So there is no way for 3 separate path components.