Let $\xi = (\pi: E \to B)$ be an $n$-dimensional vector bundle and let $u \in H^n(E, E_0; \mathbb{F}_2)$ be the mod $2$ fundamental class of $E$. (I.e. it is the class such that it is fiberwise the unique non-zero class.) Similarly to the definition of Euler classes for $\mathbb{Z}$-valued cohomology I want to understand the following
Claim. We have $\pi^*w_n(\xi) = u|_E$.
So we know by definition that $\pi^*w_k(\xi) \smile u = u \smile u$. This seems suggestive enough but some argument is missing, as we are working with two $- \smile u$'s with different domains here. (Otherwise, I would have enjoyed using that $- \smile u$ is an isomorphism.)
If $E$ were oriented, then the claim would also follow by naturality via the Euler class. (But maybe this suggests that I'm missing some argument for Euler classes which would already work for Stiefel-Whitney classes.)
How do I proceed?
We write $\pi^*\xi = (\pi' : \pi^*E \to E)$. By naturality of Stiefel-Whitney classes we have $\pi^*w_n(\xi) = w_n(\pi^*\xi)$ and so by the characteristization of Stiefel-Whitney classes, it suffices to check $$ u|_E \smile u_{\pi^*\xi} = \operatorname{Sq}^n(u_{\pi^* \xi}) = u_{\pi^*\xi}^2$$ where the last equality is since $\operatorname{rk}(\pi^*\xi) = \operatorname{rk}(\xi) = n$. On the other hand, the action of $u|_E$ on the left is really given by the action of $$\pi'^*(u|_E) = (\pi'^*u)|_{\pi^*E} = (u_{\pi^*\xi})|_{\pi^*E}. $$ The latter equality is because Thom classes pull back to Thom classes (which can be checked by restrictions on fibers). But now our desired equality is evident.