It should be clear that $H^k(B)$ corresponds to the set of all $\mathbb{S}^{k-1}$-bundles over $B$ up to isomorphism via the Euler class. For $k=2$, the addition of Euler classes should correspond to the fibrewise complex tensor product of the bundles (completet to complex line bundles).
Is there any correspondence to the cup product of Euler classes? E. g. for two Euler classes of $\mathbb{S}^1$-bundles, the cup product should be the Euler class of a $\mathbb{S}^3$-bundle. My first guess would be the fibrewise suspension of the fibrewise smash product?
Is there any good literature about this?
It is not true that $H^k(B)$ corresponds to the set of all $\mathbb S^{k-1}$ bundles up to isomorphism. For example there are $\mathbb{Z}$ non-trivial oriented $\mathbb S^{2}$ bundles on $\mathbb{S}^4$ coming from oriented three dimensional vector bundles as $\pi_3(SO(3))=\mathbb{Z}$. I use here the clutching construction to classify vector bundles over spheres, see for example Hatcher's notes.
The euler class behaves nicely with respect to direct sums over vector bundles: If $E$ and $F$ are oriented, then so is $E\oplus F$ and
$$ e(E\oplus F)=e(E)\cup e(F). $$