I will address the first version of your question (i.e. no conditions on commuting flows).
A vector bundle $E \to B$ admits $k$ linearly independent vector fields if and only if $E$ has a subbundle isomorphic to $\varepsilon^k$, the trivial rank $k$ bundle. The largest such $k$ is called the span of $E$. If $E$ has rank $n$ with span $k$, then $w_i(E) = 0$ for $i > n - k$ and $p_i(E) = 0$ for $i > \lfloor\frac{1}{2}(n-k)\rfloor$.
If $\pi : TM \to M$ denotes the natural projection, then the subbundle of $TTM$ consisting of vectors tangent to the fibers of $\pi$ is precisely $\ker(d\pi)$. Your first question can be rephrased as: what is the span of $\ker(d\pi)$? Note that $\ker(d\pi) \cong \pi^*TM$, and hence $\operatorname{span}(\ker(d\pi)) = \operatorname{span}(\pi^*TM)$. In general, $\operatorname{span}(f^*E) \geq \operatorname{span}(E)$ but in our case, $\pi : TM \to M$ is a homotopy equivalence, so we obtain $\operatorname{span}(\pi^*TM) = \operatorname{span}(TM)$ which we often call the span of $M$.
The span of a manifold is very difficult to calculate in general. For spheres, the problem was resolved by Adams in 1962 - note, the characteristic class conditions mentioned above tell us nothing here as spheres are stably parallelisable. The span of $S^{n-1}$ is $\rho(n) - 1$ where $\rho(n)$ denotes the $n^{\text{th}}$ Radon-Hurwitz number, defined as follows: if $n = 2^{4a + b}c$ where $a, b, c$ are non-negative integers, $0 \leq b \leq 3$ and $c$ is odd, then $\rho(n) = 2^b + 8a$.