ag.algebraic geometry - Grassmannian bundle theorem

Re 1.: the reasoning is correct additively, but not multiplicatively (did you check the case $k=1$?).

Re 2.: given any complex bundle $E$ over $X$ of rank $n$, the cohomology of the associated Grassmannian bundle of $k$-planes is $$H^{bullet}(X,mathbf{Z})otimesmathbf{Z}[c_1,ldots,c_k,c_1',ldots,c_{n-k}']/(1+c_1+cdots +c_k)(1+c_1'+cdots +c_{n-k}')=c(E)$$
where $deg c_i=deg c'_i=2i$ and $c(E)$ is the total Chern class of $E$.

This a modification of the Grothendieck trick which computes the cohomology of the projectivized vector bundle.

upd: here is a sketch of a proof of the above formula. On the total space $G$ of the Grassmannian vector bundle there is the tautological $k$-plane bundle $S$, which is a subbundle of the pullback $E'$ of $E$ under the bundle projection $p:Gto X$. Let $Q$ be the quotient bundle. We have $c(S)c(Q)=c(E')$. So we get an algebra map from the above algebra to the cohomology of $G$ (taking an element of $H^{bullet}(X,mathbf{Z})$ to its pullback, $c_i$ to $c_i(S)$ and $c_i'$ to $c_i(Q)$) and we have to show that it is an isomorphism.

Surjectivity: let us pick a $mathbf{Z}$-basis of the cohomology of the Grassmannian, express each element as a polynomial in the Chern classes of the tautological bundle and take the resulting polynomials in $c_i(S)$'s. When restricted to any fiber these form a $mathbf{Z}$-basis of the cohomology of the fiber, so by the Leray-Hirsch principle, $H^{bullet}(G,mathbf{Z})$ is a free $H^{bullet}(X,mathbf{Z})$-module spanned by these classes.

Injectivity: the above algebra is also a free module over $H^{bullet}(X,mathbf{Z})$ and the the above map from it to $H^{bullet}(G,mathbf{Z})$ takes a basis to a basis.