No. $\chi(S^{2n}\times S^{2n})=\chi(S^{2n})^2=4$ and so the Euler class of $S^{2n}\times S^{2n}$ is nontrivial (it pairs to the Euler characteristic under Poincare-duality). But that means there cannot exist a nonvanishing section of $TS^{2n}\oplus TS^{2n}$ (such a section would split the bundle with a trivial summand and the Euler class of a trivial bundle is zero), i.e. it is not trivial.

[[Edit]]: I hastily interpreted "sum" as the "direct product", although $\oplus$ is almost surely (or surely) referring to Whitney sum:

No. $\chi(S^{2n}\times S^{2n})=\chi(S^{2n})^2=4$ and so the Euler class of $S^{2n}\times S^{2n}$ is nontrivial (it pairs to the Euler characteristic under Poincare-duality). But that means there cannot exist a nonvanishing section of $TS^{2n}\oplus TS^{2n}$ (such a section would split the bundle with a trivial summand and the Euler class of a trivial bundle is zero), i.e. it is not trivial.