Intuition behind the Thom Isomorphism.

For a talk about Euler and Chern classes I need the Thom isomorphism with $\mathbb{Z}$ coefficients at some point (when proving the Gysin exact sequence). Recall that for an oriented real $n$-plane bundle $\xi$ with projection $\pi: E \to B$ the Thom isomorphism is given by: $$\Phi: H^k\left(B\right)\to H^{n+k}\left(E,E_0\right), \; x \mapsto\pi^{*}\left(x\right) \cup u$$

where $u$ denotes the fundamental class of $H^n \left(E,E_0\right)$. I don't have enough time to prove or even sketch the proof of this fact but I want to explain why this makes sense in 2 or 3 sentences. I'll prove the fact that $\pi$ induces an isomorphism in cohomology anyway, so I'd like to understand the cup product part of this isomorphism. So far I consulted Milnor's Characteristic classes but without success. I'd be glad if someone could give me a reference or explain this to me.

$u$ is the Thom class, it is Kronecker dual to those classes "orthogonal" to $E_0$. That is, think of those simplices $\nu$ which intersect $E_0$ in one point, then $u(\nu)=1$. Now $E_0$ is homeomorphic to $B$, so if we have a simplex $\tau$ in $B$ we can think of it is being in $E_0$. Now take $\tau\times \nu$, then the image of the $f$ where $f$ is dual to $\tau$ (that is $f(\tau)=1$) under the Thom isomorphism will be the function dual to $\tau\times \nu$, so $\Phi(f)( \tau\times \nu)=1$. Cup product comes from the external product operation $\times$, thus that is how it enters into the formula.

In Milnor, he proves Thom, via a lemma which is basically "Thom for $\mathbb{R}^n$", this is the local picture for the Thom map so I would go back and study that lemma. One can also try to analyse the theorem in terms of Poincare duality, in which case you are taking an intersection with $E_0$ and then dualising.

There is a very nice geometric explanation for the cup product part by the de Rham theory.

Let $B$ be an oriented $n$-manifold($\mathbb{Z}$-coefficients). First recall that in de Rham cohomology, the cup product corresponds to the wedge product of differential forms. For oriented vector bundles there is another cohomology called compact vertical cohomology. Let $p: E \to B$ be an oriented vector bundle. Then cochain groups for this cohomology are given by $$ \Omega_{cv}^n(E)=\{ \omega \in \Omega^n (E) \ | \ p^{-1}(K) \cap \text{Supp}(\omega) \text{ is compact, } \forall \text{ compact } K \subset B\} $$ In particular, $\text{Supp}(\omega|_{p^{-1}(x)})$ is compact. Choose a trivialization for $E$ with coordinates $(t_1,\ldots,t_n)$ in the fibers and define the integration along the fiber as $$ p_{\ast} : \Omega_{cv}^{\ast} (E) \to \Omega^{\ast -n}(B) \\ (p^{\ast}\phi) f(x,t_1,\ldots,t_n)dt_1 \ldots dt_n \mapsto \phi \int_{\mathbb{R}^n} f(x,t_1,\ldots,t_n)dt_1 \ldots dt_n \ \text{ where } \phi \in \Omega^{\ast-n}(B)$$ and as $0$ for different types of forms. This gives an isomorphism $\Phi^{-1}$ in the question. Then Thom class $u$ is given by $u = \Phi(1)$ where $1 \in H^0(B)$(can think of bump form in each fiber) and $\Phi$ is exactly given by $\eta \mapsto p^{\ast}(\eta) \wedge u$.

More Geometry: Let $M \subset B$ be closed oriented $m$-dim submanifold. We have a natural $(n-m)$-vector bundle which is the normal bundle $\nu(M)$. Then the Thom class of the normal bundle is the Poincare dual of $S$ and Thom isomorphism relates cohomology of $M$ with the relative cohomology of the tubular neighborhood of $M$ in $B$.

Now as Rene Schipperus pointed out in his answer one can see that intersection of oriented submanifolds corresponds to wedge product.