Choose a diagram as in Lemma \ref{lemma-morphism-ringed-topoi-comes-from-morphism-ringed-sites}. Recall that being a flat module is intrinsic (see Section \ref{section-intrinsic} and Definition \ref{definition-flat}). Hence it suffices to prove the lemma for the morphism $(h, h^\sharp) : (\Sh(\mathcal{C}'), \mathcal{O}_{\mathcal{C}'}) \to (\Sh(\mathcal{D}'), \mathcal{O}_{\mathcal{D}'})$. In other words, we may assume that our sites $\mathcal{C}$ and $\mathcal{D}$ have all finite limits and that $f$ is a morphism of sites induced by a continuous functor $u : \mathcal{D} \to \mathcal{C}$ which commutes with finite limits. \medskip\noindent Recall that $f^*\mathcal{F} = \mathcal{O}_\mathcal{C} \otimes_{f^{-1}\mathcal{O}_\mathcal{D}} f^{-1}\mathcal{F}$ (Definition \ref{definition-pushforward}). By Lemma \ref{lemma-flat-change-of-rings} it suffices to prove that $f^{-1}\mathcal{F}$ is a flat $f^{-1}\mathcal{O}_\mathcal{D}$-module. Combined with the previous paragraph this reduces us to the situation of the next paragraph. \medskip\noindent Assume $\mathcal{C}$ and $\mathcal{D}$ are sites which have all finite limits and that $u : \mathcal{D} \to \mathcal{C}$ is a continuous functor which commutes with finite limits. Let $\mathcal{O}$ be a sheaf of rings on $\mathcal{D}$ and let $\mathcal{F}$ be a flat $\mathcal{O}$-module. Then $u$ defines a morphism of sites $f : \mathcal{C} \to \mathcal{D}$ (Sites, Proposition \ref{sites-proposition-get-morphism}). To show: $f^{-1}\mathcal{F}$ is a flat $f^{-1}\mathcal{O}$-module. Let $U$ be an object of $\mathcal{C}$ and let $$ f^{-1}\mathcal{O}|_U \xrightarrow{(f_1, \ldots, f_n)} f^{-1}\mathcal{O}|_U^{\oplus n} \xrightarrow{(s_1, \ldots, s_n)} f^{-1}\mathcal{F}|_U $$ be a complex of $f^{-1}\mathcal{O}|_U$-modules. Our goal is to construct a factorization of $(s_1, \ldots, s_n)$ on the members of a covering of $U$ as in Lemma \ref{lemma-flat-eq} part (2). Consider the elements $s_a \in f^{-1}\mathcal{F}(U)$ and $f_a \in f^{-1}\mathcal{O}(U)$. Since $f^{-1}\mathcal{F}$, resp.\ $f^{-1}\mathcal{O}$ is the sheafification of $u_p\mathcal{F}$ we may, after replacing $U$ by the members of a covering, assume that $s_a$ is the image of an element $s'_a \in u_p\mathcal{F}(U)$ and $f_a$ is the image of an element $f'_a \in u_p\mathcal{O}(U)$. Then after another replacement of $U$ by the members of a covering we may assume that $\sum f'_as'_a$ is zero in $u_p\mathcal{F}(U)$. Recall that the category $(\mathcal{I}_U^u)^{opp}$ is directed (Sites, Lemma \ref{sites-lemma-directed}) and that $u_p\mathcal{F}(U) = \colim_{(\mathcal{I}_U^u)^{opp}} \mathcal{F}(V)$ and $u_p\mathcal{O}(U) = \colim_{(\mathcal{I}_U^u)^{opp}} \mathcal{O}(V)$. Hence we may assume there is a pair $(V, \phi) \in \Ob(\mathcal{I}_U^u)$ where $V$ is an object of $\mathcal{D}$ and $\phi$ is a morphism $\phi : U \to u(V)$ of $\mathcal{D}$ and elements $s''_a \in \mathcal{F}(V)$ and $f''_a \in \mathcal{O}(V)$ whose images in $u_p\mathcal{F}(U)$ and $u_p\mathcal{O}(U)$ are equal to $s'_a$ and $f'_a$ and such that $\sum f''_a s''_a = 0$ in $\mathcal{F}(V)$. Then we obtain a complex $$ \mathcal{O}|_V \xrightarrow{(f''_1, \ldots, f''_n)} \mathcal{O}|_V^{\oplus n} \xrightarrow{(s''_1, \ldots, s''_n)} \mathcal{F}|_V $$ and we can apply the other direction of Lemma \ref{lemma-flat-eq} to see there exists a covering $\{V_i \to V\}$ of $\mathcal{D}$ and for each $i$ a factorization $$ \mathcal{O}|_{V_i}^{\oplus n} \xrightarrow{B''_i} \mathcal{O}|_{V_i}^{\oplus l_i} \xrightarrow{(t''_{i1}, \ldots, t''_{il_i})} \mathcal{F}|_{V_i} $$ of $(s''_1, \ldots, s''_n)|_{V_i}$ such that $B_i \circ (f''_1, \ldots, f''_n)|_{V_i} = 0$. Set $U_i = U \times_{\phi, u(V)} u(V_i)$, denote $B_i \in \text{Mat}(l_i \times n, f^{-1}\mathcal{O}(U_i))$ the image of $B''_i$, and denote $t_{ij} \in f^{-1}\mathcal{F}(U_i)$ the image of $t''_{ij}$. Then we get a factorization $$ f^{-1}\mathcal{O}|_{U_i}^{\oplus n} \xrightarrow{B_i} f^{-1}\mathcal{O}|_{U_i}^{\oplus l_i} \xrightarrow{(t_{i1}, \ldots, t_{il_i})} \mathcal{F}|_{U_i} $$ of $(s_1, \ldots, s_n)|_{U_i}$ such that $B_i \circ (f_1, \ldots, f_n)|_{U_i} = 0$. This finishes the proof.