Let $X$ be a scheme over $S$ and let $\{X_i \to X\}_{i \in I}$ be a $\tau$-covering. Let $s_i \in F'(X_i)$ be elements such that $s_i$ and $s_j$ map to the same element of $F'(X_i \times_X X_j)$ for all $i, j \in I$. We have to show that there is a unique element $s \in F'(X)$ restricting to $s_i \in F'(X_i)$ for all $i \in I$. \medskip\noindent Special case: $X$ is an affine such that the structure morphism maps into an affine open $U$ of $S$ and the covering $\{X_i \to X\}_{i \in I}$ is a standard $\tau$-covering. In this case we can write $$ X = \lim V_k $$ as a cofiltered limit with $V_k \to U$ of finite presentation and $V_k$ affine. See Algebra, Lemma \ref{algebra-lemma-ring-colimit-fp}. By Lemma \ref{lemma-limit-fppf-topology} there exists a $k$ and a standard $\tau$-covering $\{V_{k, i} \to V_k\}_{i \in I}$ whose base change to $X$ is the given covering. For $k' \geq k$ denote $\{V_{k', i} \to V_{k'}\}_{i \in I}$ the base change to $V_{k'}$ of our covering. Then we see that \begin{align*} F'(X) & = \colim_{k' \geq k} F(V_k) \\ & = \colim_{k' \geq k} \text{Equalizer}( \xymatrix{ \prod F(V_{k', i}) \ar@<1ex>[r] \ar@<-1ex>[r] & \prod F(V_{k', i} \times_{V_{k'}} V_{k', j}) } \\ & = \text{Equalizer}( \xymatrix{ \colim_{k' \geq k} \prod F(V_{k', i}) \ar@<1ex>[r] \ar@<-1ex>[r] & \colim_{k' \geq k} \prod F(V_{k', i} \times_{V_{k'}} V_{k', j}) } \\ & = \text{Equalizer}( \xymatrix{ \prod F'(X_i) \ar@<1ex>[r] \ar@<-1ex>[r] & \prod F'(X_i \times_X X_j) } \end{align*} The first equality holds by construction of $F'$. The second holds by assumption (c). The third holds because filtered colimits are exact. The fourth again holds by construction of $F'$. In this way we find that the sheaf property holds for $F'$ with respect to $\{X_i \to X\}_{i \in I}$. \medskip\noindent General case. Choose an affine open covering $X = \bigcup U_k$ such that each $U_k$ maps into an affine open of $S$. For every $k$ we may choose a standard $\tau$-covering $\{V_{k, j} \to U_k\}_{j = 1, \ldots, m_k}$ which refines $\{X_i \times_X U_k \to U_k\}_{i \in I}$. For each $j \in \{1, \ldots, m_k\}$ choose an index $i_{k, j} \in I$ and a morphism $g_{k, j} : V_{k, j} \to X_{i_{k, j}}$ over $X$. Let $s_{k, j}$ be the element of $F'(V_{k, j})$ we get by restricting $s_{i_{k, j}}$ via $g_{k, j}$. Observe that $s_{k, j}$ and $s_{k', j'}$ restrict to the same element of $F'(V_{k, j} \times_X V_{k', j'})$ for all $k$ and $k'$ and all $j \in \{1, \ldots, m_k\}$ and $j' \in \{1, \ldots, m_{k'}\}$; verification omitted. In particular, by the result of the previous paragraph there is a unique element $s_k \in F'(U_k)$ restricting to $s_{k, j}$ for all $j$. With this notation we are ready to finish the proof. \medskip\noindent Proof of uniqueness of $s$: this is true because $F'$ satisfies the sheaf property for Zariski coverings and $s|_{U_k}$ must be equal to $s_k$ because both restrict to $s_{k, j}$ for all $j$. This uniqueness then shows that $s_k$ and $s_{k'}$ must restrict to the same section of $F'$ over (the non-affine scheme) $U_k \cap U_{k'}$ because these sections restrict to the same section over the $\tau$-covering $\{V_{k, j} \times_X V_{k', j'} \to U_k \cap U_{k'}\}$. Thus by the sheaf property for Zariski coverings, there is a unique section $s$ of $F'$ over $X$ whose restriction to $U_k$ is $s_k$. We omit the verification (similar to the above) that $s$ restricts to $s_i$ over $X_i$.