Daniel Litt's answer is completely correct, of course. You can also see the result using projective geometry (perhaps this is how Chow proves the result). For every $1$-dimensional subspace $L$ of $F^{\oplus n}$, there is an induced embedding, $$\phi_L:\text{Grass}(k-1,F^{\oplus n}/L) \hookrightarrow \text{Grass}(k,F^{\oplus n}),$$ obtained by associating to every subspace $\overline{V}$ of $F^{\oplus n}/L$ the inverse image $V$ in $F^{\oplus n}$. Denote by $P(t)$ the Hilbert polynomial of $\text{Image}(\phi_L)$ with respect to the ample invertible sheaf $\omega^\vee$, where $\omega$ is the dualizing sheaf of $\text{Grass}(k,F^{\oplus n})$. Then there is a well-defined morphism, $$\Phi:\text{Grass}(1,F^{\oplus n}) \to \text{Hilb}^{P(t)}_{\text{Grass}(k,F^{\oplus n})}, \ \ [L]\mapsto [\text{Image}(\phi_L)].$$ I claim that $\Phi$ gives an isomorphism of the domain with one of the connected components of the target (when $n\neq 2k$, there is only one connected component, so that $\Phi$ is an $F$-isomorphism). Moreover, it is equivariant for the natural actions of $\textbf{PGL}_n$. Therefore, for every central simple algebra $A$ of "degree" $n$, there is an induced isomorphism of $F$-schemes, $$ SB_1(A) \to \text{Hilb}^{P(t)}_{SB_k(A)},$$ where again the Hilbert polynomial is with respect to the very ample invertible sheaf $\omega^\vee$ on $SB_k(A)$.

In particular, if $SB_k(A)$ is isomorphic to $\text{Grass}(k,F^{\oplus n})$, then every connected component of $\text{Hilb}^{P(t)}_{SB_k(A)}$ has an $F$-rational point. Thus, also $SB_1(A)$ has an $F$-rational point. That implies that $A$ is split by the result quoted by the OP.

Of course one still needs to verify that $\Phi$ is an isomorphism. However, since this is a purely geometric result, I imagine that most of us are happier with this claim than the original problem. Also it is fairly easy to prove using some "homogeneity" and cohomology vanishing.