The Thom class and Thom isomorphism theorem for oriented vector bundles are proven ( at least to my knowledge) by induction on the open covers and some manipulation with Mayer-Vietoris sequences.


What is the "actual reason" behind the existence of Thom class? It seems strange that such an interesting class would exist just because some Mayer-Vietoris sequences routinely produce it.



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.


