Is it possible to give an example of $n$ dimensional manifold with the property that the tangent bundle $TM$ cannot be expressed as Whitney sum of two subbundles? It is certain true for two sphere; it is certainly not true for three dimensional manifold since every three manifold is parallelizable. You can always split $TM$ as $\gamma \oplus Q$ where $\gamma$ is one dimensional if you can find non vanishing vector field (i.e. where the Euler class is zero which is the case for odd dimension) but what about examples with $n$ even, larger than $2$?



Splitting of tangent bundle
To expand on my comment above, suppose that $TS^{2k}\cong\xi\oplus\eta$ for some non-trivial vector bundles $\xi$ and $\eta$ over $S^{2k}$ of dimensions $m$ and $\ell$, respectively. Hence $0<m,\ell<2k$. Since $S^{2k}$ is simply-connected, both $\xi$ and $\eta$ are oriented, hence possess Euler classes
$$
e(\xi)\in H^m(S^{2k};\mathbb{Z}),\qquad e(\eta)\in H^{\ell}(S^{2k};\mathbb{Z}).
$$
Since $S^{2k}$ is $(2k-1)$-connected, both are zero.


However, Property 9.6 of Euler classes in Milnor and Stasheff then gives that that
$$e(TS^{2k})=e(\xi)\cup e(\eta)=0.$$
This is a contradiction, since
$$
0\neq 2=\chi(S^{2k}) = \langle e(TS^{2k}),[S^{2k}]\rangle
$$
(this is Corollary 11.12 in Milnor and Stasheff).


