How could people classify all rank $2$ complex vector bundles over $S^2\times S^2$ up to isomorphism? Could you give a rank 2 complex vector bundle which cannot be split as a sum of two line bundles?
This is a special case of The space of homotopy classes of maps of products of spheres.
Classifying rank 2 complex vector bundles on $S^2\times S^2$ is the same as calculating the set of pointed homotopy classes $\langle S^2\times S^2,BU(2)\rangle$. For this I would use the cofibration sequence $$ S^3\to S^2\vee S^2 \to S^2\times S^2 \to S^4\to S^3\vee S^3\to \cdots $$ where the first map is the attaching map of the top cell of $S^2\times S^2$, and the fourth map is its suspension, therefore is nullhomotopic. The map $q:S^2\times S^2\to S^4$ can be identified with collapsing the complement of a small open ball $B^4\subset S^2\times S^2$.
Taking maps into $BU(2)$ results in an exact sequence $$ 0 \to \pi_4(BU(2))\to \langle S^2\times S^2,BU(2)\rangle \to \pi_2(BU(2))\oplus \pi_2(BU(2))\to \pi_3(BU(2)) $$ which reduces (using $\pi_i(U(2))=\mathbb{Z},0,\mathbb{Z}$ for $i=1,2,3$) to a short exact sequence of sets $$ 0\to \mathbb{Z} \to \langle S^2\times S^2, BU(2)\rangle \to \mathbb{Z}\oplus\mathbb{Z} \to 0. $$ This is not quite the full classification (which I'm sure must appear in the literature somewhere, if you look hard enough), but allows us to say some things.
For instance, the second map is restriction to either $S^2$ factor, so I think this shows us how to produce a rank 2 complex bundle over $S^2\times S^2$ which is not a product of line bundles: take a nontrivial rank 2 bundle over $S^4$ (such as are classified by their $c_2$, by Greg Arone's answer to your previous question), and pull it back via the collapse map $q:S^2\times S^2\to S^4$.

$\begingroup$ Does the sequence mean the homotopic class $[S^2\times S^2,U(2)]$ has a group structure? If so, how to define the group structure? Thanks $\endgroup$ – DLIN Aug 9 '16 at 4:58

$\begingroup$ Well, the set $\langle S^2\times S^2, U(2)\rangle$ certainly has a group structure (since $U(2)$ is an Hspace), but I do not know if $\langle S^2\times S^2, BU(2)\rangle$ admits a group structure, because $BU(2)$ is not an Hspace. $\endgroup$ – Mark Grant Aug 9 '16 at 7:50