If $L/K$ is a Galois extension, then one can define a site in which the objects are intermediate fields $K \subseteq E \subseteq L$ which are finite over $K$. Or, you can take finite étale algebras $A$ over $K$ with a homorphism of $K$-algebras $A \to L$. The topology is the étale topology. These two sites define the same topos. There is an equivalence between sheaves on either of these sites and discrete $\mathrm{Gal}(L/K)$-modules, hence cohomology coincides with Galois cohomology.
Does this answer your question? I am not sure I understand what you have in mind.
The following is a counterexample for $\mathcal{F}$ a presheaf of abelian groups (or sets, if you like).
Let $X=\lbrace a,b,c,d\rbrace$ with nontrival opens given by $\lbrace a \rbrace,\lbrace b \rbrace,U=\lbrace a,b,c \rbrace,V=\lbrace a,b,d \rbrace, U\cap V$.
Define the presheaf $\mathcal{F}$ by
$\mathcal{F}(\lbrace a \rbrace)=\mathcal{F}(\lbrace b \rbrace)=\mathbb{Z}/2\mathbb{Z}$,
$\mathcal{F}(U)=\mathcal{F}(V)=\mathcal{F}(U\cap V)=\mathcal{F}(X)=\mathbb{Z}$,
with the obvious restriction maps.
Then $\mathcal{F}^+(X)=\lbrace (x,y)\in\mathbb{Z}\oplus\mathbb{Z}| x\equiv y\text{ (mod 2)}\rbrace$, since the germs at $a$ and $b$ are determined by those at $c$ and $d$, and the only restrictions on $c$ and $d$ are that they give the same germs at $a$ and $b$.
Consider $(0,2)\in\mathcal{F}^+(X)$. The germs at $c$ and $d$ cannot come from a common section of $\mathcal{F}(X)$. Any system of sections which does not include $X$ in the cover must include both $U$ and $V$, having sections 0 and 2, respectively. Of course these do not agree when restricted to $U\cap V$. QED
This construction relies crucially on the fact that the presheaf is not separated (i.e. gluing is not unique). If the presheaf $\it{is}$ separated, the condition described in the question is clearly satisfied.
This construction was shown to me by Paul Balmer.
Best Answer
It's not diagonal. A section of $\mathcal F$ on $U$ has a germ at each element of $\text{Hom}_{X}(\bar{x},U)$, and you take the product of all those germs, which may be different.
It's not so hard to see that if you tried to do it diagonally, it wouldn't be functorial for automorphisms of $U$, say in the case when $U$ is a finite étale cover.