Set $F = G(k, n)$. To prove the lemma we will use the criterion of Schemes, Lemma \ref{schemes-lemma-glue-functors}. The reason $F$ satisfies the sheaf property for the Zariski topology is that we can glue sheaves, see Sheaves, Section \ref{sheaves-section-glueing-sheaves} (some details omitted). \medskip\noindent The family of subfunctors $F_i$. Let $I$ be the set of subsets of $\{1, \ldots, n\}$ of cardinality $n - k$. Given a scheme $S$ and $j \in \{1, \ldots, n\}$ we denote $e_j$ the global section $$ e_j = (0, \ldots, 0, 1, 0, \ldots, 0)\quad(1\text{ in }j\text{th spot}) $$ of $\mathcal{O}_S^{\oplus n}$. Of course these sections freely generate $\mathcal{O}_S^{\oplus n}$. Similarly, for $j \in \{1, \ldots, n - k\}$ we denote $f_j$ the global section of $\mathcal{O}_S^{\oplus n - k}$ which is zero in all summands except the $j$th where we put a $1$. For $i \in I$ we let $$ s_i : \mathcal{O}_S^{\oplus n - k} \longrightarrow \mathcal{O}_S^{\oplus n} $$ which is the direct sum of the coprojections $\mathcal{O}_S \to \mathcal{O}_S^{\oplus n}$ corresponding to elements of $i$. More precisely, if $i = \{i_1, \ldots, i_{n - k}\}$ with $i_1 < i_2 < \ldots < i_{n - k}$ then $s_i$ maps $f_j$ to $e_{i_j}$ for $j \in \{1, \ldots, n - k\}$. With this notation we can set $$ F_i(S) = \{q : \mathcal{O}_S^{\oplus n} \to \mathcal{Q} \in F(S) \mid q \circ s_i \text{ is surjective}\} \subset F(S) $$ Given a morphism $f : T \to S$ of schemes the pullback $f^*s_i$ is the corresponding map over $T$. Since $f^*$ is right exact (Modules, Lemma \ref{modules-lemma-exactness-pushforward-pullback}) we conclude that $F_i$ is a subfunctor of $F$. \medskip\noindent Representability of $F_i$. To prove this we may assume (after renumbering) that $i = \{1, \ldots, n - k\}$. This means $s_i$ is the inclusion of the first $n - k$ summands. Observe that if $q \circ s_i$ is surjective, then $q \circ s_i$ is an isomorphism as a surjective map between finite locally free modules of the same rank (Modules, Lemma \ref{modules-lemma-map-finite-locally-free}). Thus if $q : \mathcal{O}_S^{\oplus n} \to \mathcal{Q}$ is an element of $F_i(S)$, then we can use $q \circ s_i$ to identify $\mathcal{Q}$ with $\mathcal{O}_S^{\oplus n - k}$. After doing so we obtain $$ q : \mathcal{O}_S^{\oplus n} \longrightarrow \mathcal{O}_S^{\oplus n - k} $$ mapping $e_j$ to $f_j$ (notation as above) for $j = 1, \ldots, n - k$. To determine $q$ completely we have to fix the images $q(e_{n - k + 1}), \ldots, q(e_n)$ in $\Gamma(S, \mathcal{O}_S^{\oplus n - k})$. It follows that $F_i$ is isomorphic to the functor $$ S \longmapsto \prod\nolimits_{j = n - k + 1, \ldots, n} \Gamma(S,  \mathcal{O}_S^{\oplus n - k}) $$ This functor is isomorphic to the $k(n - k)$-fold self product of the functor $S \mapsto \Gamma(S, \mathcal{O}_S)$. By Schemes, Example \ref{schemes-example-global-sections} the latter is representable by $\mathbf{A}^1_\mathbf{Z}$. It follows $F_i$ is representable by $\mathbf{A}^{k(n - k)}_\mathbf{Z}$ since fibred product over $\Spec(\mathbf{Z})$ is the product in the category of schemes. \medskip\noindent The inclusion $F_i \subset F$ is representable by open immersions. Let $S$ be a scheme and let $q : \mathcal{O}_S^{\oplus n} \to \mathcal{Q}$ be an element of $F(S)$. By Modules, Lemma \ref{modules-lemma-finite-type-surjective-on-stalk}. the set $U_i = \{s \in S \mid (q \circ s_i)_s\text{ surjective}\}$ is open in $S$. Since $\mathcal{O}_{S, s}$ is a local ring and $\mathcal{Q}_s$ a finite $\mathcal{O}_{S, s}$-module by Nakayama's lemma (Algebra, Lemma \ref{algebra-lemma-NAK}) we have $$ s \in U_i \Leftrightarrow \left( \text{the map } \kappa(s)^{\oplus n - k} \to \mathcal{Q}_s/\mathfrak m_s\mathcal{Q}_s \text{ induced by } (q \circ s_i)_s \text{ is surjective} \right) $$ Let $f : T \to S$ be a morphism of schemes and let $t \in T$ be a point mapping to $s \in S$. We have $(f^*\mathcal{Q})_t = \mathcal{Q}_s \otimes_{\mathcal{O}_{S, s}} \mathcal{O}_{T, t}$ (Sheaves, Lemma \ref{sheaves-lemma-stalk-pullback-modules}) and so on. Thus the map $$ \kappa(t)^{\oplus n - k} \to (f^*\mathcal{Q})_t/\mathfrak m_t(f^*\mathcal{Q})_t $$ induced by $(f^*q \circ f^*s_i)_t$ is the base change of the map $\kappa(s)^{\oplus n - k} \to \mathcal{Q}_s/\mathfrak m_s\mathcal{Q}_s$ above by the field extension $\kappa(s) \subset \kappa(t)$. It follows that $s \in U_i$ if and only if $t$ is in the corresponding open for $f^*q$. In particular $T \to S$ factors through $U_i$ if and only if $f^*q \in F_i(T)$ as desired. \medskip\noindent The collection $F_i$, $i \in I$ covers $F$. Let $q : \mathcal{O}_S^{\oplus n} \to \mathcal{Q}$ be an element of $F(S)$. We have to show that for every point $s$ of $S$ there exists an $i \in I$ such that $s_i$ is surjective in a neighbourhood of $s$. Thus we have to show that one of the compositions $$ \kappa(s)^{\oplus n - k} \xrightarrow{s_i} \kappa(s)^{\oplus n} \rightarrow \mathcal{Q}_s/\mathfrak m_s\mathcal{Q}_s $$ is surjective (see previous paragraph). As $\mathcal{Q}_s/\mathfrak m_s\mathcal{Q}_s$ is a vector space of dimension $n - k$ this follows from the theory of vector spaces.