Intuition behind the Thom Isomorphism.

$u$ 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.