Let $\mathcal F\xrightarrow{\alpha}\mathcal G\xrightarrow{\beta}\mathcal H$ be an exact sequence of abelian sheaves on the topological space $X$, so that $\operatorname{ker}\beta\cong \operatorname{im}\alpha$ as sheaves (in particular they are subpresheaves of $\mathcal G$). Then for every point $x\in X$, also $(\operatorname{ker}\beta)_x\cong (\operatorname{im}\alpha)_x$.
We want to prove that also the sequence of abelian groups $\mathcal F_x\xrightarrow{\alpha_x}\mathcal G_x\xrightarrow{\beta_x}\mathcal H_x$ is exact, i.e. that $\operatorname{ker}\beta_x\cong \operatorname{im}\alpha_x$; my idea would be to prove that $(\operatorname{ker}\beta)_x\cong \operatorname{ker}\beta_x$ and $ (\operatorname{im}\alpha)_x \cong \operatorname{im}\alpha_x$.
Start with $(\operatorname{ker}\beta)_x\cong \operatorname{ker}\beta_x$. Define a map $(\operatorname{ker}\beta)_x\to \operatorname{ker}\beta_x$: if $s_x\in (\operatorname{ker}\beta)_x$ is the stalk at $x$ of $s\in \operatorname{ker}\beta(U)$, for an open set $x\in U\subseteq X$, we send it to the stalk $s_x\in \mathcal G_x$ of $s$ viewed as an element of $\mathcal G(U)$. It's obvious that such map doesn't depend on the choice of $s$ and is also a homomorphism. Plus $\beta_x(s_x)=(\beta (s))_x$ by definition, and $\beta (s)=0$, so we constructed a homomorphism $(\operatorname{ker}\beta)_x\to \operatorname{ker}\beta_x$. Now we need the inverse, so take $s_x\in \mathcal G_x$, with $s\in \mathcal G(U)$, such that $\beta_x(s_x)=0$, i.e. $\beta(s)|_V=0$ for some open set $x\in V\subseteq U$. Then $s|_V\in \operatorname{ker}\beta$, so that we can send $s_x$ to $(s|_V)_x$. This map doesn't depend on the choice of $V$ nor on the choice of $s$, and it is indeed an inverse of our homomorphism, that hence is bijective.
For the other equality the situation is simpler: since by definition the images the stalks are the stalks of the images, the map $\operatorname{im}\alpha_x\to (\operatorname{im}\alpha)_x:\alpha_x(r_x)\mapsto (\alpha(U)(r))_x$, for $r\in \mathcal F(U)$ with an open set $x\in U\subseteq X$, is automatically a bijective homomorphism.
At this point, knowing that $\operatorname{ker}\beta\cong \operatorname{im}\alpha$, we have $(\operatorname{ker}\beta)_x\cong (\operatorname{im}\alpha)_x$, and so $\operatorname{ker}\beta_x\cong \operatorname{im}\alpha_x$, meaning that passing to the stalks preserves the exactness of sequences. Now consider the exact sequence $0\to \mathcal F'\to \mathcal F\to \mathcal F/\mathcal F'\to 0$, where $\mathcal F,\mathcal F'$ are sheaves, that yields an exact sequence $0\to \mathcal F'_x\to \mathcal F_x\to (\mathcal F/\mathcal F')_x\to 0$, so that $(\mathcal F/\mathcal F')_x\cong \mathcal F_x/\mathcal F'_x$. When considering the sheafication $(\mathcal F/\mathcal F')^\dagger$, we also have $(\mathcal F/\mathcal F')^\dagger_x\cong \mathcal F_x/\mathcal F'_x$. Thus I'd say that, starting from $0\to \mathcal F'\to \mathcal F\to \mathcal F/\mathcal F'\to 0$ as above, one can only say that $0\to \mathcal F'\to \mathcal F\to (\mathcal F/\mathcal F')^\dagger$ is exact, since the canonical morphism $\mathcal F/\mathcal F'\to(\mathcal F/\mathcal F')^\dagger$ is injective, while $0\to \mathcal F'_x\to \mathcal F_x\to (\mathcal F/\mathcal F')^\dagger _x\to 0$ stays exact.
Do yo agree with my conclusions? I'm still not familiar with sheaves so I'm a bit confused.
If $$0\to\mathcal{F}'\to\mathcal{F}\to\mathcal{F}/\mathcal{F}'\to 0$$ is exact as a sequence of presheaves, then $$0\to\mathcal{F}'_x\to\mathcal{F}_x\to(\mathcal{F}/\mathcal{F}')_x\to 0$$ is exact as a sequence of abelian group, and since $(\mathcal{F}/\mathcal{F}')_x=(\mathcal{F}/\mathcal{F}')^\dagger_x$, and since this holds for every $x$, the sequence $$0\to\mathcal{F}'\to\mathcal{F}\to(\mathcal{F}/\mathcal{F}')^\dagger\to 0$$ is exact as a sequence of sheaves. (In particular, $0\to\mathcal{F}'\to\mathcal{F}\to(\mathcal{F}/\mathcal{F}')^\dagger$ is exact as a sequence of presheaves, and the last map is not onto in the category of presheaves).
Note that