Can I comb unoriented hair on a ball?

An alternative way to see why this isn't true is to fix a Riemannian metric. If such a subbundle existed, then restricting to those vectors of norm $1$ would give a two-sheeted covering of $S^2$. If it were disconnected, there would be a nowhere zero vector field on $S^2$, a contradiction. If it were connected, $S^2$ would admit a connected two-sheeted covering, a contradiction with it being simply connected.

No. Line bundles on $S^2$ are classified by $H^1(S^2;\mathbb{Z}/2)=0$ which means that there is only the trivial line bundle $L$ on $S^2$. But then $L\oplus L$ is the trivial bundle hence not the tangent bundle of $S^2$.

An interesting question is if $TS^n$ splits as a direct sum $E\oplus F$ of low dimensional bundles. If $n$ is odd the Euler characteristic is zero, so there is a non-zero vector field. So let us assume $n$ is even.

If $E$ and $F$ are required to be orientable bundles this is not possible: The Euler class is multiplicative and $e(TS^n)$ is twice the generator of $H^n(S^n;\mathbb{Z})$ if $n$ is even and in particular it is non-zero. But if $E$ and $F$ are proper subbundles the euler classes $e(E)$ and $e(F)$ lie in some $H^i(S^n;\mathbb{Z})$ for $0<i<n$ hence $e(E)\cup e(F)=0$ which contradicts the fact that $e(E)\cup e(F)=e(TS^n)\not=0$.

A bundle is orientable iff its determinant bundle has a section, i.e. is a trivial line bundle. But we have seen that the sphere only admits the trivial line bundle. This implies that every bundle on $S^n$ is orientable and the previous argument shows that the tangent bundle cannot split at all.

This discussion btw does not imply that there are no non-trivial bundles on $S^n$ of rank $r<n$, and indeed there are some.