I’ll write the weight of $S^\vee$ as $(0,\ldots,0,-1)$; it might be helpful to think of this bundle as $\bigwedge^{k-1}S \otimes \det(S)^{-1}$. For $\mathrm{Sym}^2(S)$ the weight is $(2,0,\ldots,0)$; both of these are vectors with $k$ entries. By the Pieri rule,
$$S^\vee \otimes \mathrm{Sym}^2(S) \cong S \oplus \mathbb{S}_{(2,0,\ldots,0,-1)}(S),$$ where $\mathbb{S}$ denotes the Schur functor.
Similarly the weight of $Q^\vee$ is $(0,\ldots,0,-1)$ (with $n-k$ entries). So we concatenate the weights for each of the two summands:
$$w = (0,\ldots,0,-1,1,0,\ldots,0) \text{ and } w'= (0,\ldots,0,-1,2,0,\ldots, 0,-1).$$
By Borel-Weil-Bott we add $\rho = (n,n-1,\ldots,1)$ and count inversions in the resulting word; if there is a repeat, all the cohomology vanishes.
In the first case there is exactly one inversion, so the bundle has nonvanishing $H^1$. After sorting, the resulting weight is
$$\mathrm{sort}(w+\rho)-\rho = \vec{0}$$ So $H^1$ is one-dimensional (the trivial representation). For the other, $w'+\rho$ has a repeat from the $2$ and the $0$ two steps before it. So all the cohomology vanishes.
Edit: Exception: for the second calculation, if $n-k=1$ then the repeat doesn’t occur. In that case I guess there is one inversion and nonvanishing $H^1$ of weight $(1,0,\ldots,0,-1)$ which has rank $n^2-1$; it is the kernel of $V \otimes V^\vee \to \mathbb{C}$, i.e. the traceless matrices. Also, if $k=1$ then the second sum and simply doesn’t occur at all. I think the calculation above is correct otherwise, that is if $n-k$ and $k$ are both $\geq 2$.