Let $F(M)$ be this space of vector fields. Let's work this out for an even-dimensional sphere $M=S^n=S^{2p}$, so $n=2p$ and $k=2p-1$. I claim that it is not rationally $3$-connected, if $n\ge 3$.
This space $F(S^{n})$ fibers over the space $Conf_2(S^{n})\sim \mathbb RP^{n}$ of unordered pairs of points in $S^n$, by recording where the field vanishes. Call the fiber (over, say, some pair of antipodal points) $F_0(S^{n})$. Then $F_0(S^n)$ fibers over $GL_n^+(\mathbb R)\times Gl_n^+(\mathbb R)\sim SO_n\times SO_n$ by recording the derivative of the field at those two points. The fiber of $F_0(S^n)\to O_n\times O_n$ is homotopy equivalent to the space of all nowhere-vanishing tangent fields on $S^{n-1}\times I$ having some specified restriction to the boundary, and that's the same as the space of all paths in $Map(S^{n-1},S^{n-1})$ between a certain pair of points. So $F(S^n)$ fibers over $A=\mathbb RP^n$ and the fiber is fibered over $B=SO_n\times SO_n$ and the fiber of that is $C=\Omega Map(S^{n-1},S^{n-1})$.
Now $A$ is rationally $(n-1)$-connected, and $C$ is rationally $(n-2)$-connected, but $B$ has rationally nontrivial $\pi_3$ if $n\ge 3$, so the same is true of $F(S^{2p})$.
In fact, I think that looking more closely we will find that it is not even $1$-connected.