Let $k$ be an algebraically closed field, $X=\mathbb{P}^1$ the projective line. Let $P=(1:0)$ and $Q=(0:1)$ be points on $X$, and $\mathscr{F}$ be the sheaf of regular functions on $X$. Define a subsheaf $\mathscr{F}'$ of $\mathscr{F}$ by $$\mathscr{F}'(U) = \left\{ \begin{align} &\{f \in \mathscr{F}(U); f(P)=f(Q)=0\}, &\text{if } P,Q \in U \\ &\{f \in \mathscr{F}(U); f(P)=0\}, & \text{if } P\in U, Q\notin U \\ &\{f \in \mathscr{F}(U); f(Q)=0\}, & \text{if } Q\in U, P\notin U \\ &\mathscr{F}(U), &\text{if } Q\notin U \text{and } P\notin U \end{align} \right.$$ where $U$ is any open set in $X$. Define $\mathscr{F}'' = \mathscr{F}/\mathscr{F}'$.
The point here is that $$0\longrightarrow \mathscr{F}' \longrightarrow \mathscr{F} \longrightarrow \mathscr{F}'' \longrightarrow 0$$ gives a short exact sequence, but the natural map $$ \mathscr{F}(X) \longrightarrow \mathscr{F}''(X)$$ is (supposedly) not surjective.
Note that $\mathscr{F}(X) = k,$ and $\mathscr{F}'(X) = \{0\}$. Still, I fail to describe the global sections of $\mathscr{F}''$; can somebody help out?
The quotient sheaf $\mathcal{F}'' = \mathcal{F} / \mathcal{F}'$ is defined to be the sheafification of the presheaf where the sections are $\mathcal{F}(U)/ \mathcal{F}'(U)$. Since the global sections of closed projective variety are contant, the global sections on the quotient presheaf are just $k$. However, take $U = Y\setminus \{P\}$ and $f \in \mathcal{F}''(U)$ (as an element of the presheaf) to be the constant $f = a \in k$ and similarly $V = Y\setminus \{Q\}$ with $g \in \mathcal{F}''(V)$ the constant $b \in k $ $ ( a\neq b)$. These sections agree at their intersection (since $U \cap V = Y\setminus \{P,Q\}$ and $\mathcal{F}''(U\cap V) = 0)$ and thus can be glued together into a non-constant section on $Y$. In this way $F''(Y)$ (this time as the associated sheaf) $\cong k \times k$