Intuition behind Thom class

It is easy to understand the existence of a Thom class by considering cellular cohomology. Let the given vector bundle be $E\to B$ with fibers of dimension $n$. One can assume without significant loss of generality that $B$ is a CW complex with a single 0-cell. The Thom space $T(E)$ is the quotient $D(E)/S(E)$ of the unit disk bundle of $E$ by the unit sphere bundle. One can give $T(E)$ a CW structure with $S(E)/S(E)$ as the only 0-cell and with an $(n+k)$-cell for each $k$-cell of $B$. These cells in $T(E)$ arise from pulling back the bundle $D(E)\to B$ via characteristic maps $D^k\to B$ for the $k$-cells of $B$. These pullback are products since $D^k$ is contractible.

In particular, $T(E)$ has a single $n$-cell and an $(n+1)$-cell for each 1-cell of $B$. There are no cells in $T(E)$ between dimension $0$ and $n$. The cellular boundary of an $(n+1)$-cell is 0 if $E$ is orientable over the corresponding 1-cell of $B$, and it is twice the $n$-cell in the opposite case. Thus $H^n(T(E);{\mathbb Z})$ is $\mathbb Z$ if $E$ is orientable and $0$ if $E$ is non-orientable. In the orientable case a generator of $H^n(T(E);{\mathbb Z})$ restricts to a generator of $H^n(S^n;{\mathbb Z})$ in the "fiber" $S^n$ of $T(E)$ over the 0-cell of $B$, hence the same is true for all the "fibers" $S^n$ and so one has a Thom class.

One not-very technical way to think of the Thom Isomorphism Theorem is that if you have a vector bundle, $p : E \to B$, if you remove the $0$-section $Z$ of the vector bundle from the Thom space $Th(p)$, you get a contractible space. So given a homology class in $H_* Th(p)$ the obstruction to trivializing it can be thought of as its intersection with $Z$. If there's no intersection, you're in the contractible space $Th(p) \setminus Z$. So the intersection of a homology class with $Z$ is tautologically the thing that keeps track of the homology class itself.

That's how I like to think of the Thom Isomorphism Theorem. So why is there a Thom class? Because you can intersect with $Z$. In cohomology this is cupping with the Thom class since that's what intersections translate to in cohomology.

You are thinking in terms of ordinary cohomology, where Mayer-Vietoris patches together the always present local orientation to produce a global one when you have it. It is more advanced, but maybe more illuminating, to note that the definition in general is intrinsically global. An $n$-plane bundle $p$ over a space $B$ has an associated sphere bundle $Sph(p)$ (by fiberwise one point compactification) with based fibers and thus a section. The quotient $Sph/B$ is the Thom space $T$ of $p$. For a multiplicative cohomology theory $E$, a Thom class $\mu$ is an element of $\tilde{E}^n(T)$ whose restriction to $\tilde{E}^n(S^n_b)\cong \tilde{E}^0(S^0)$ is a unit in this ring for any $b\in B$, where $S^n_b$ is the fiber over $b$ in $Sph(p)$. This definition is admitttedly mysterious. It suffices to give a Thom isomorphism and it is important geometrically, but the real explanation is more advanced and still not very well known. One should think of $E^*$ as represented by a ring spectrum $E$. Bundle theory naturally concerns spaces over $B$, or parametrized spaces. One can make sense of parametrized spectra over $B$, and one can even take the smash product of a parametrized space and a spectrum to obtain a parametrized spectrum. Thus one can make sense of $Sph(p)\wedge E$ as a spectrum over $B$. Of course, there is also a trivial spherical bundle $B\times S^n$ over $B$. It turns out that a Thom class as I defined it cohomologically is the same thing as a trivialization: an equivalence of parametrized spectra between $Sph(p)\wedge E$ and $(B\times S^n)\wedge E$. That is the geometric meaning. This is proven in the book Parametrized Homotopy Theory, by Sigurdsson and myself (available on my website).