I am reading Edwin Spanier's book "Algebraic Topology", theorem 5.8.19.
I do not understand the very first sentence of the proof. I hope there is not too much notational confusion here, but the context is that $\dot{p} : \dot{E}\to B$ is a $q$-sphere bundle, and $E$ is the mapping cone of $\dot{p}$, which is also fibered over $B$, $p : E\to B$, as a $q+1$-disk bundle in a natural way, containing $\dot{E}$ as a sub-bundle. $R$ is a PID, and $U$ is any element of $H^{q+1}(E,\dot{E};R)$ which restricts, in any fiber $(p^{-1}(b), \dot{p}^{-1}(b))$, to a generator of the fiber cohomology $H^{q+1}(p^{-1}(b), \dot{p}^{-1}(b);R)$ (generates as an $R$-module.)
Spanier has proven earlier that, to each path $\omega : b_0\to b_1$ in the base space $B$, we can choose a lift $h[\omega]$, which is a continuous map $p^{-1}(b_0) \to p^{-1}(b_1)$. This choice is unique up to homotopy, so it determines a functor from the fundamental groupoid of $B$ into the homotopy category of topological spaces, sending each point to the fiber above it, and each path to $h[\omega]$. Here $h[\omega]^\ast$ is just the pullback on cohomology induced by the map.
My question is then why the equality should hold. I do not exactly see why it should be true, even (as Spanier implies) in the case where we restrict to an open set on which we can trivialize the fiber bundle. A minor point perhaps but I am a bit pedantic about details.
I'll suppose you are able to verify it for the case $B=[0,1]$ and the path given by the identity. Your path $\omega$ gives us a sphere bundle over $[0,1]$, $\Gamma$, via pullback. Your construction then gives a map from the relative cohomology of the fiber over 0 to the relative cohomology of the fiber over 1 which you will see sends the restriction of the orientation to the restriction of the orientation.
Now we may pick the orientation of $\Gamma$ to be the pullback of the orientation of $\xi$. This gives us a commuting diagram with vertical maps pullbacks and horizontal maps $h[\omega]$ and $h[\operatorname{Id}]$. Because $h[Id]$ sends restriction of the orientation class to restriction of the orientation class, and the diagram is commutative, then it must be true of $h[\omega]$.