The proof works slightly differently; the $G_0$ one needs may be not the one you name but a retract of it. In other words, the map you ask about may be nonsurjective but it has a retraction.
The assumption can be equivalently formulated as follows: for any $\alpha:h_S\to B$, the pullback of $i:A\hookrightarrow B$ along it is a representable subfunctor $h_Y\hookrightarrow h_S$. In particular, as you noted, the pullback of $i$ along $\Psi:h_G\to B$ is a representable subfunctor $h_M\hookrightarrow h_G$. Now let us further pull this subfunctor back along $\Phi:A\to h_G$. Since $\Psi\circ\Phi=i$, what we obtain is isomorphic to the pullback of $i$ along itself, i. e. to $A$: $$ \require{AMScd} \begin{CD} A@<<< h_M@<<<A\\ @ViVV @VVV @|\\ B @<<\Psi<h_G@<<\Phi<A \end{CD}, $$ and the upper horizontal composite is the identity of $A$. Thus although $A$ might be not isomorphic to $h_M$, it is a retract of it, hence representable itself by some $G_0$ (in fact by the kernel/image/cokernel of the idempotent $M\to M$ corresponding by Yoneda to the composite $h_M\to A\to h_M$).
PS
By the way it follows (if my argument is correct) that it suffices to restrict the above assumption to $\Psi:h_G\to B$ only rather than arbitrary $\alpha:h_S\to B$. However if one looks at the place where Mumford verifies this assumption in the situation he needs it, it is maybe even easier to work with arbitrary $S$ rather than use the specific properties of $G$ (the latter is certain Grassmanian in his situation).