The first plus construction is separated and preserves stalks

Let $\sigma ,\tau \in {\mathcal{F}}^{+}(U)$ and suppose there is an open cover $U={\bigcup }_{\alpha \in A}{W}_{\alpha }$ such that $\sigma {\mathrm{\mid }}_{W_{\alpha }}=\tau {\mathrm{\mid }}_{W_{\alpha }}$ for every $\alpha$. Fix $x\in U$, and choose $\alpha$ with $x\in W_{\alpha }$. Equality on $W_{\alpha }$ means that the two restricted presentations determine the same germ at $x$, so $\sigma$ and $\tau$ have the same germ at $x$. Since this holds for every $x\in U$, the two global presentations are equivalent by [F1]. Therefore $\sigma =\tau$, and ${\mathcal{F}}^{+}$ is separated by [F2].

To prove surjectivity of ${\eta }_x$, let $\xi \in ({\mathcal{F}}^{+}{)}_x$. Choose an open neighbourhood $U$ of $x$ and a section $\sigma \in {\mathcal{F}}^{+}(U)$ representing $\xi$. Pick a local presentation $(U_i,s_i)$ of $\sigma$ with $x\in U_i$ for some index $i$. Then on the neighbourhood $U_i$ the section $\sigma$ equals ${\eta }_{U_i}(s_i)$ by [F1], so the germ $\xi$ is the image of the germ of $s_i$ at $x$. Hence ${\eta }_x$ is surjective.

Now suppose ${\eta }_x(a)={\eta }_x(b)$ in $({\mathcal{F}}^{+}{)}_x$, where $a\in {\mathcal{F}}_x$ and $b\in {\mathcal{F}}_x$. Represent $a$ and $b$ by sections $s\in \mathcal{F}(U)$ and $t\in \mathcal{F}(V)$ on neighbourhoods of $x$. By [L1], equality of their images in the stalk of ${\mathcal{F}}^{+}$ means that there exists an open neighbourhood $W\subseteq U\cap V$ of $x$ such that ${\eta }_W(s{\mathrm{\mid }}_W)={\eta }_W(t{\mathrm{\mid }}_W)\text{ in }{\mathcal{F}}^{+}(W)\mathrm{.}$ By [F1], equality of these two single-chart classes says exactly that $s{\mathrm{\mid }}_W$ and $t{\mathrm{\mid }}_W$ have the same germ at every point of $W$, in particular at $x$. Thus $a=b$ in ${\mathcal{F}}_x$, so ${\eta }_x$ is injective.

Steps 1.2 and 1.3 prove that every ${\eta }_x$ is bijective, and step 1.1 proves that ${\mathcal{F}}^{+}$ is separated.