= Solution
Let $n=\operatorname{rank}_{\mathbb Z}M$. Work with <commutative rings> and take a quotient to mean a surjection $p:M\otimes_{\mathbb Z}R\twoheadrightarrow Q$, considered up to an <isomorphism> of $Q$ commuting with $p$. Equivalently, retain its <kernel> as a submodule of $M\otimes R$. Here a rank-$r$ <locally free module> is finite locally free, or equivalently a <finite projective module> of constant rank $r$. A <ring homomorphism> $R\to R'$ sends the quotient to
$$
M\otimes R'\twoheadrightarrow Q\otimes_RR'.
$$
The <tensor product of modules> is right exact and preserves finite projectivity, so this defines the functorial action. We prove <scheme> representability in the usual <functor of points> sense:
$$
\boxed{F(R)\cong\operatorname{Hom}_{\mathbb Z}
\bigl(\operatorname{Spec}R,\operatorname{Gr}_r(M)\bigr),}
$$
naturally in $R$. The <Grassmannian of locally free quotients> will be constructed explicitly.
Assume first $0<r<n$, and choose a <basis> $e_1,\ldots,e_n$ of $M$. For every ordered subset $I=\{i_1<\cdots<i_r\}\subseteq\{1,\ldots,n\}$, consider those quotients for which $p(e_{i_1}),\ldots,p(e_{i_r})$ are a <basis> of $Q$. This choice gives a unique identification $Q\cong R^r$, in which the quotient is represented by an $r\times n$ <matrix> $A_I$ whose columns indexed by $I$ are the identity <matrix>. Its remaining $r(n-r)$ entries are arbitrary elements of $R$. Thus this subfunctor is represented by the <affine scheme>
$$
U_I=\operatorname{Spec}\mathbb Z[a_{kj}:1\leq k\leq r,\ j\notin I]
\cong\mathbb A^{r(n-r)}_{\mathbb Z}.
$$
There is no further change-of-basis ambiguity: the selected quotient images already specify its <basis> uniquely.
For another subset $J$, write $A_I^{(J)}$ for the square submatrix consisting of the $J$ columns. The same quotient lies in the $J$ chart exactly when its <determinant> is a unit. Consequently the chart overlap is the <principal open subset>
$$
U_{IJ}=D(\delta_{IJ})\subseteq U_I,\qquad
\delta_{IJ}=\det A_I^{(J)}.
$$
On that overlap the <matrix> normalized in the $J$ <basis> is
$$
\boxed{A_J=(A_I^{(J)})^{-1}A_I.}
$$
The entries are regular on $D(\delta_{IJ})$ by the <adjugate matrix>. Conversely, the $I$ columns of $A_J$ form $(A_I^{(J)})^{-1}$, so normalization back to $I$ recovers $A_I$. The transition is therefore an <isomorphism> $U_{IJ}\cong U_{JI}$.
The cocycle identity follows from the same <matrix> computation, not just an assertion about <bases>. On a triple overlap, normalization first to $J$ and then to $K$ gives
$$
\begin{aligned}
(A_J^{(K)})^{-1}A_J
&=\left((A_I^{(J)})^{-1}A_I^{(K)}\right)^{-1}
(A_I^{(J)})^{-1}A_I\\
&=(A_I^{(K)})^{-1}A_I.
\end{aligned}
$$
This is direct normalization from $I$ to $K$. By <gluing of schemes along open subschemes>, the <standard affine charts of the quotient Grassmannian> therefore glue to a <scheme> $G$ over $\mathbb Z$.
The universal <matrices> also glue a universal quotient. On $U_I$, use the surjection
$$
M\otimes_{\mathbb Z}\mathcal O_{U_I}\xrightarrow{\ A_I\ }
\mathcal O_{U_I}^{\oplus r}.
$$
On overlaps, identify its targets by $(A_I^{(J)})^{-1}$, precisely the transition <matrix> above. The cocycle identity glues these free sheaves into a rank-$r$ <locally free sheaf> $\mathcal Q$ and glues the maps into $M\otimes\mathcal O_G\twoheadrightarrow\mathcal Q$. Its <kernel> $\mathcal S$ is locally free of rank $n-r$: on $U_I$ an explicit <basis> is
$$
e_j-\sum_{k=1}^ra_{kj}e_{i_k},\qquad j\notin I.
$$
Hence we have the universal <exact sequence>
$$
\boxed{0\longrightarrow\mathcal S\longrightarrow
M\otimes_{\mathbb Z}\mathcal O_G\longrightarrow\mathcal Q
\longrightarrow0,\qquad \operatorname{rank}\mathcal S=n-r,
\quad\operatorname{rank}\mathcal Q=r.}
$$
It remains to verify that these charts capture every quotient, including a nonfree projective target. Given $p:R^n\twoheadrightarrow Q$, define $V_I\subseteq\operatorname{Spec}R$ as the locus where the selected images $p(e_i)$, $i\in I$, form a <basis>. This is open: locally trivialize the <locally free sheaf> associated with $Q$, and it is the nonvanishing locus of the corresponding maximal minor. The sets $V_I$ cover $\operatorname{Spec}R$. Indeed, at every prime $\mathfrak p$, the images of all $n$ vectors span the $r$-dimensional residue-field quotient, so some $r$ of them form a <basis> there. Their <determinant> is a unit in $R_{\mathfrak p}$, hence they form a <basis> over that <local ring> and on an open neighbourhood of $\mathfrak p$.
On $V_I$, trivialize $Q$ using exactly those selected images. The remaining columns give regular functions and hence a morphism $V_I\to U_I$. On $V_I\cap V_J$, the normalized <matrices> satisfy the displayed transition formula, so the local morphisms glue to a morphism
$$
f_p:\operatorname{Spec}R\longrightarrow G.
$$
Its pullback of the universal quotient is the original quotient, because this is true in each of the specified <bases>. Conversely, any morphism $f:\operatorname{Spec}R\to G$ pulls back the universal sequence to a rank-$r$ locally free quotient. On each inverse-image chart its <matrix> is exactly the normalized <matrix> used to reconstruct $f$, so the two constructions are inverse. They are unchanged by an <isomorphism> commuting with the quotient map, and they commute with every <base change> $R\to R'$. This proves the natural bijection and therefore
$$
\boxed{G=\operatorname{Gr}_r(M)\text{ represents }F.}
$$
Taking kernels over a field identifies this with the usual <Grassmannian> of $(n-r)$-dimensional subspaces of $M\otimes k$. The construction itself does not assume the target <module> is globally free.
For $r=0$, the only quotient is zero; for $r=n$, a surjection to a <locally free module> of the same rank is an <isomorphism>, as can be checked locally by its <determinant>. Both endpoint functors are represented by $\operatorname{Spec}\mathbb Z$. If $r>n$, no such quotient exists over a nonempty test <scheme>, and the representing <scheme> is empty. Finally, <scheme> representability does not mean representation by a single <ring>: already $r=1,n=2$ gives $\mathbb P^1_{\mathbb Z}$, which is not affine. This is the standard algebraic-geometric meaning of representability for the functor on <rings> here.
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