Choose a finite type ring map $A \to C$ and a finite $C$-module $N$ and a prime $\mathfrak q$ of $C$ such that $B = C_{\mathfrak q}$ and $M = N_{\mathfrak q}$. In the following, when we say ``the theorem holds for $(N/C/A, \mathfrak q)$ we mean that it holds for $(A \to B, M)$ where $B = C_{\mathfrak q}$ and $M = N_{\mathfrak q}$. By Lemma \ref{lemma-flat-at-point-go-up} the functor $F_{lf}$ is unchanged if we replace $B$ by a local ring flat over $B$. Hence, since $A$ is henselian, we may apply Lemma \ref{lemma-existence-algebra} and assume that there exists a complete d\'evissage of $N/C/A$ at $\mathfrak q$. \medskip\noindent Let $(A_i, B_i, M_i, \alpha_i, \mathfrak q_i)_{i = 1, \ldots, n}$ be such a complete d\'evissage of $N/C/A$ at $\mathfrak q$. Let $\mathfrak q'_i \subset A_i$ be the unique prime lying over $\mathfrak q_i \subset B_i$ as in Definition \ref{definition-complete-devissage-at-x-algebra}. Since $C \to A_1$ is surjective and $N \cong M_1$ as $C$-modules, we see by Lemma \ref{lemma-flat-at-point-finite} it suffices to prove the theorem holds for $(M_1/A_1/A, \mathfrak q'_1)$. Since $B_1 \to A_1$ is finite and $\mathfrak q_1$ is the only prime of $B_1$ over $\mathfrak q'_1$ we see that $(A_1)_{\mathfrak q'_1} \to (B_1)_{\mathfrak q_1}$ is finite (see Algebra, Lemma \ref{algebra-lemma-unique-prime-over-localize-below} or More on Morphisms, Lemma \ref{more-morphisms-lemma-finite-morphism-single-point-in-fibre}). Hence by Lemma \ref{lemma-flat-at-point-finite} it suffices to prove the theorem holds for $(M_1/B_1/A, \mathfrak q_1)$. \medskip\noindent At this point we may assume, by induction on the length $n$ of the d\'evissage, that the theorem holds for $(M_2/B_2/A, \mathfrak q_2)$. (If $n = 1$, then $M_2 = 0$ which is flat over $A$.) Reversing the last couple of steps of the previous paragraph, using that $M_2 \cong \Coker(\alpha_2)$ as $B_1$-modules, we see that the theorem holds for $(\Coker(\alpha_1)/B_1/A, \mathfrak q_1)$. \medskip\noindent Let $A'$ be an object of $\mathcal{C}$. At this point we use Lemma \ref{lemma-induction-step} to see that if $(M_1 \otimes_A A')_{\mathfrak q'}$ is flat over $A'$ for a prime $\mathfrak q'$ of $B_1 \otimes_A A'$ lying over $\mathfrak m_{A'}$, then $(\Coker(\alpha_1) \otimes_A A')_{\mathfrak q'}$ is flat over $A'$. Hence we conclude that $F_{lf}$ is a subfunctor of the functor $F'_{lf}$ associated to the module $\Coker(\alpha_1)_{\mathfrak q_1}$ over $(B_1)_{\mathfrak q_1}$. By the previous paragraph we know $F'_{lf}$ is corepresented by $A/J$ for some ideal $J \subset A$. Hence we may replace $A$ by $A/J$ and assume that $\Coker(\alpha_1)_{\mathfrak q_1}$ is flat over $A$. \medskip\noindent Since $\Coker(\alpha_1)$ is a $B_1$-module for which there exist a complete d\'evissage of $N_1/B_1/A$ at $\mathfrak q_1$ and since $\Coker(\alpha_1)_{\mathfrak q_1}$ is flat over $A$ by Lemma \ref{lemma-complete-devissage-flat-finite-type-module} we see that $\Coker(\alpha_1)$ is free as an $A$-module, in particular flat as an $A$-module. Hence Lemma \ref{lemma-induction-step} implies $F_{lf}(A')$ is nonempty if and only if $\alpha \otimes 1_{A'}$ is injective. Let $N_1 = \Im(\alpha_1) \subset M_1$ so that we have exact sequences $$ 0 \to N_1 \to M_1 \to \Coker(\alpha_1) \to 0 \quad\text{and}\quad B_1^{\oplus r_1} \to N_1 \to 0 $$ The flatness of $\Coker(\alpha_1)$ implies the first sequence is universally exact (see Algebra, Lemma \ref{algebra-lemma-flat-universally-injective}). Hence $\alpha \otimes 1_{A'}$ is injective if and only if $B_1^{\oplus r_1} \otimes_A A' \to N_1 \otimes_A A'$ is an isomorphism. Finally, Theorem \ref{theorem-flattening-map} applies to show this functor is corepresentable by $A/I$ for some ideal $I$ and we conclude $F_{lf}$ is corepresentable by $A/I$ also. \medskip\noindent To prove the final statement, suppose that $A \to B$ is essentially of finite presentation and $M$ of finite presentation over $B$. Let $I \subset A$ be the ideal such that $F_{lf}$ is corepresented by $A/I$. Write $I = \bigcup I_\lambda$ where $I_\lambda$ ranges over the finitely generated ideals contained in $I$. Then, since $F_{lf}(A/I) = \{*\}$ we see that $F_{lf}(A/I_\lambda) = \{*\}$ for some $\lambda$, see Lemma \ref{lemma-flat-at-point} part (2). Clearly this implies that $I = I_\lambda$.