Assume (1). Let $\mathcal{I} \subset \mathcal{O}_U$ be the sheaf of ideals generated by $f_1, \ldots, f_n$. Then $\sum f_j \otimes s_j$ is a section of $\mathcal{I} \otimes_{\mathcal{O}_U} \mathcal{F}|_U$ which maps to zero in $\mathcal{F}|_U$. As $\mathcal{F}|_U$ is flat (Lemma \ref{lemma-restriction-flat}) the map $\mathcal{I} \otimes_{\mathcal{O}_U} \mathcal{F}|_U \to \mathcal{F}|_U$ is injective. Since $\mathcal{I} \otimes_{\mathcal{O}_U} \mathcal{F}|_U$ is the sheaf associated to the presheaf tensor product, we see there exists a covering $\{U_i \to U\}$ such that $\sum f_j|_{U_i} \otimes s_j|_{U_i}$ is zero in $\mathcal{I}(U_i) \otimes_{\mathcal{O}(U_i)} \mathcal{F}(U_i)$. Unwinding the definitions using Algebra, Lemma \ref{algebra-lemma-relations} we find $t_{i1}, \ldots, t_{i l_i} \in \mathcal{F}(U_i)$ and $a_{ijk} \in \mathcal{O}(U_i)$ such that $\sum_j a_{ijk}f_j|_{U_i} = 0$ and $s_j|_{U_i} = \sum_k a_{ijk}t_{ik}$. Thus (2) holds. \medskip\noindent Assume (2). Let $U$, $n$, $m$, $A$ and $s_1, \ldots, s_n$ as in (3) be given. Observe that $A$ has $m$ columns. We will prove the assertion of (3) is true by induction on $m$. For the base case $m = 0$ we can use the factorization through the zero sheaf (in other words $l_i = 0$). Let $(f_1, \ldots, f_n)$ be the last column of $A$ and apply (2). This gives new diagrams $$ \mathcal{O}_{U_i}^{\oplus m} \xrightarrow{B_i \circ A|_{U_i}} \mathcal{O}_{U_i}^{\oplus l_i} \xrightarrow{(t_{i1}, \ldots, t_{il_i})} \mathcal{F}|_{U_i} $$ but the first column of $A_i = B_i \circ A|_{U_i}$ is zero. Hence we can apply the induction hypothesis to $U_i$, $l_i$, $m - 1$, the matrix consisting of the first $m - 1$ columns of $A_i$, and $t_{i1}, \ldots, t_{il_i}$ to get coverings $\{U_{ij} \to U_j\}$ and factorizations $$ \mathcal{O}_{U_{ij}}^{\oplus l_i} \xrightarrow{C_{ij}} \mathcal{O}_{U_{ij}}^{\oplus k_{ij}} \xrightarrow{(v_{ij1}, \ldots, v_{ij k_{ij}})} \mathcal{F}|_{U_{ij}} $$ of $(t_{i1}, \ldots, t_{il_i})|_{U_{ij}}$ such that $C_i \circ B_i|_{U_{ij}} \circ A|_{U_{ij}} = 0$. Then $\{U_{ij} \to U\}$ is a covering and we get the desired factorizations using $B_{ij} = C_i \circ B_i|_{U_{ij}}$ and $v_{ija}$. In this way we see that (2) implies (3). \medskip\noindent Assume (3). Let $\mathcal{G} \to \mathcal{H}$ be an injective homomorphism of $\mathcal{O}$-modules. We have to show that $\mathcal{G} \otimes_\mathcal{O} \mathcal{F} \to \mathcal{H} \otimes_\mathcal{O} \mathcal{F}$ is injective. Let $U$ be an object of $\mathcal{C}$ and let $s \in (\mathcal{G} \otimes_\mathcal{O} \mathcal{F})(U)$ be a section which maps to zero in $\mathcal{H} \otimes_\mathcal{O} \mathcal{F}$. We have to show that $s$ is zero. Since $\mathcal{G} \otimes_\mathcal{O} \mathcal{F}$ is a sheaf, it suffices to find a covering $\{U_i \to U\}_{i \in I}$ of $\mathcal{C}$ such that $s|_{U_i}$ is zero for all $i \in I$. Hence we may always replace $U$ by the members of a covering. In particular, since $\mathcal{G} \otimes_\mathcal{O} \mathcal{F}$ is the sheafification of $\mathcal{G} \otimes_{p, \mathcal{O}} \mathcal{F}$ we may assume that $s$ is the image of $s' \in  \mathcal{G}(U) \otimes_{\mathcal{O}(U)} \mathcal{F}(U)$. Arguing similarly for $\mathcal{H} \otimes_\mathcal{O} \mathcal{F}$ we may assume that $s'$ maps to zero in $\mathcal{H}(U) \otimes_{\mathcal{O}(U)} \mathcal{F}(U)$. Write $\mathcal{F}(U) = \colim M_\alpha$ as a filtered colimit of finitely presented $\mathcal{O}(U)$-modules $M_\alpha$ (Algebra, Lemma \ref{algebra-lemma-module-colimit-fp}). Since tensor product commutes with filtered colimits (Algebra, Lemma \ref{algebra-lemma-tensor-products-commute-with-limits}) we can choose an $\alpha$ such that $s'$ comes from some $s'' \in \mathcal{G}(U) \otimes_{\mathcal{O}(U)} M_\alpha$ and such that $s''$ maps to zero in $\mathcal{H}(U) \otimes_{\mathcal{O}(U)} M_\alpha$. Fix $\alpha$ and $s''$. Choose a presentation $$ \mathcal{O}(U)^{\oplus m} \xrightarrow{A} \mathcal{O}(U)^{\oplus n} \to M_\alpha \to 0 $$ We apply (3) to the corresponding complex of $\mathcal{O}_U$-modules $$ \mathcal{O}_U^{\oplus m} \xrightarrow{A} \mathcal{O}_U^{\oplus n} \xrightarrow{(s_1, \ldots, s_n)} \mathcal{F}|_U $$ After replacing $U$ by the members of the covering $U_i$ we find that the map $$ M_\alpha \to \mathcal{F}(U) $$ factors through a free module $\mathcal{O}(U)^{\oplus l}$ for some $l$. Since $\mathcal{G}(U) \to \mathcal{H}(U)$ is injective we conclude that $$ \mathcal{G}(U) \otimes_{\mathcal{O}(U)} \mathcal{O}(U)^{\oplus l} \to \mathcal{H}(U) \otimes_{\mathcal{O}(U)} \mathcal{O}(U)^{\oplus l} $$ is injective too. Hence as $s''$ maps to zero in the module on the right, it also maps to zero in the module on the left, i.e., $s$ is zero as desired.