After the definition of orientation for a vector bundle$^{(1)}$ in "Characteristic Classes" by Milnor/Stasheff (page 96), the author make this comment:
The local compatibility condition implies that for every point in the base space there exists a neighborhood $N$ and a cohomology class $$u \in H^n (\pi^{-1}(N), \pi^{-1}(N)_0;Z)$$ so that for every fiber $F$ over $N$ the restriction $$u | (F, F_0) \in H^n(F, F_0; Z)$$ is equal to $u_F$.
Where $u_F$ is the generator associated to the orientation of the fiber F.
I understand the intuition behind this comment, but I don't see exactly where this $u$ comes from. My idea is that an orientation in $\mathbb{R}^n$ gives a class in $H^n(\mathbb{R}^n, \mathbb{R}^n_0; Z)$ and this should give a class in $H^n(N \times \mathbb{R}^n, N \times \mathbb{R}^n_0; Z) \cong H^n (\pi^{-1}(N), \pi^{-1}(N)_0;Z)$ somehow.
$^{(1)}$ The definition in the book is:
An orientation for $\xi$ is a function which assigns an orientation to each fiber $F$ of $\xi$, subjected to the following local compatibility condition. For every point $b_0$ in the base space there should exist a local coordinate system $(N, h)$, with $b_0 \in N$ and $h: N \times \mathbb{R}^n \to \pi^{-1}(N)$, so that for each fiber $F = \pi^{-1}(b)$ over N the homomorphism $x \mapsto h(b, x)$ from $\mathbb{R}^n$ to F is orientation preserving.
As the cohomology of $(\mathbb{R}^n, \mathbb{R}^n_0)$ is torsion-free, and in particular, $$ H^q( \mathbb{R}^n, \mathbb{R}^n_0) = \begin{cases} \mathbb{Z} & q = n \\ 0 & q \neq n \end{cases},$$
by the Kunneth formula $$ H^m(N \times \mathbb{R}^n, N \times \mathbb{R}^n_0) \cong \oplus_{p+q=m} H^p(N) \otimes H^q(\mathbb{R}^n,\mathbb{R}^n_0) = H^{m-n}(N).$$ If $N$ is connected, the class you seek in $H^n(N \times \mathbb{R^n},N \times \mathbb{R}^n_0)$ is exactly the image of the generator in $H^0(N)$. Geometrically, the Kunneth isomorphism tells us this class in $H^n(N \times \mathbb{R}^n, N \times \mathbb{R}^n_0)$ is exactly a generator of $H^n(\mathbb{R}^n,\mathbb{R}^n_0)$ over every point in $N$.