If $n$ is odd then $S^{n-1}$ doesn't admit a nowhere-vanishing vector field, and if $n$ is even then there does exist one (Hairy Ball Theorem). We can then ask, on $S^{n-1}$, what is the maximum number $k(n)$ of linearly independent vector fields? Rewriting $n=2^{4a+b}(2s+1)$, Adams computes $k(n)=2^b+8a-1$.

In particular, on the $(4n+1)$-spheres, there is only one nowhere-vanishing vector field up to linear-independence, whereas in every other (odd) dimension there are more.
Example: on the circle $S^1$ there are the vector fields generated by (counter)clockwise rotation, but these are the same up to a scalar. This makes sense: I start flowing along this single dimension and then I have to continue flowing in that direction until I come back to my starting point.
I tried considering the difference between $S^3$ and $S^5$, which fiber over $\mathbb{C}P^1$ and $\mathbb{C}P^2$ respectively. A nowhere-vanishing vector field in both cases is given by taking the standard nowhere-vanishing vector field on the $S^1$-fiber. But for $S^3$ there are three linearly-independent fields (the $i,j,k$-directions when representing $S^3$ as the unit quaternions -- is there a way to see this using the fibration picture?), whereas for some reason $S^5$ can only admit the one.

What can be the differential/topological reasoning behind this? I.e. is there a down-to-earth way to deduce this result on $S^{4n+1}$, or for starters, $S^5$?

Could there possibly be an analogous index theorem going on here, in the same way that the Poincare-Hopf theorem provides us the Hairy Ball result?