I am surely late to his, but maybe it has some worth noticing the following statement: If $M$ is a closed manifold of dimension $2n$ such that its odd integer cohmology vanishes, then every complex rank $n$ vector bundle over $M$ is uniquely determined up to isomorphism by its Chern classes.
One explanation is provided though $K$-theory. Since we are in the stable range the set of isomorphism classes of complex vector bundles of rank $n$ $\mathrm{Vect(M)}$ is in bijection to the reduced complex $K$-theory of $M$, $\widetilde K(M)$. Moreover $\widetilde K(M)$ has no torsion since $H^\ast(M,\mathbb Z)$ has no torsion and the Chern character $\mathrm{ch}\colon \widetilde K(M) \to \widetilde H^\ast(M,\mathbb Q)$ is injective. This implies that every stable class of complex vector bundle is uniquely determined by its Chern character, hence by its Chern classes (since, again, integer cohomology has no torsion). This also induces a group structure on $\mathrm{Vect}(M)$ by the group structure of $\widetilde K(M)$.
In that case a complex vector bundle $E \to S^2\times S^2$ of rank $2$ is a Whitney sum of two line bundles, say $L_1$ and $L_2$ with first Chern classes $x$ and $y$, then $c_1(E) = x+y$ and $c_2(E)=x\cup y$. If $p \colon S^2 \times S^2 \to S^4$ is a map of degree $1$ then, pulling back the tautological bundle of $S^4 = \mathbb{HP}^1$ to $S^2\times S^2$ via $p$, we obtain a bundle which cannot split into the sum of two line bundles for the reasons mentioned above.