Left exactness of $-(U)$ A few questions about the proof given here: $\Gamma(U,\cdot)$ is a left-exact functor $\mathfrak{Ab}(X)\to\mathfrak{Ab}$

Let $0\to\mathcal{F}'\xrightarrow{\phi}\mathcal{F}\xrightarrow{\psi}\mathcal{F}''$ be a left-exact sequence. Then for any open set $U\subseteq X$, the sequence $0\to\Gamma(U,\mathcal{F}')\to\Gamma(U,\mathcal{F})\to\Gamma(U,\mathcal{F}'')$ is left-exact.


$$\begin{array}{cccccc}
0 & \rightarrow & \Gamma(U,\mathcal{F}') & \xrightarrow{\phi_U} & \Gamma(U,\mathcal{F}) & \xrightarrow{\psi_U} & \Gamma(U,\mathcal{F}'')\\
& & \downarrow & & \downarrow & & \downarrow\\
0 & \rightarrow & \mathcal{F}'_p & \xrightarrow{\phi_p} & \mathcal{F}_p & \xrightarrow{\psi_p} & \mathcal{F}^{''}_p
\end{array}$$


Now if $s\in\Gamma (U,\mathcal{F}')$, we have $\psi_U(\phi_U(s))_p=\psi_p(\phi_p(s_p))=0$ for every $p\in U$, hence $\psi_U(\phi_U(s))=0$. The inclusion left to show is $\ker(\psi_U)\subseteq\operatorname{im}(\phi_U)$.

First, I suppose vertical maps are denoted by $(-)_p$. Also, I suppose the LHS of $\psi_U(\phi_U(s))_p=\psi_p(\phi_p(s_p))$ should be $(\psi_U(\phi_U(s)))_p$ (the image of $\psi_U(\phi_U(s))$ under the vertical arrow). I don't see how $(\psi_U(\phi_U(s)))_p=0$ in $\mathcal F''_p$ implies $\psi_U(\phi_U(s))=0$ in $\mathcal F''(U)$. The former means that there is an equality of equivalence classes $[(\psi_U(\phi_U(s)), U)]=[(0,X)]$, which only means that $res_{U,W}(\psi_U(\phi_U(s)))=res_{X,W}(0)$ for some $W\subset U$. How does this imply that $\psi_U(\phi_U(s))=0$?
Next, the answer there says:

In my opinion, you need to reason why the $s'_P$ do glue to some $s'\in\mathcal{F}'(U)$. However, this is not that tough. We pick neighborhoods $V_P\subseteq U$ for each point $P$ such that $s'_P=(f_P,V_P)$ with $f_P\in\mathcal{F}'(V_P)$. For $W:=V_P\cap V_Q\ne 0$, we have $\phi_W(f_P|_W)=t|_W=\phi_W(f_Q|_W)$ and we know that $\phi_W$ is injective.

Here, I suppose, the notation $f_P|_W $ means $res_{V_P,W}(f_P)$ and similarly for $f_Q|_W$ and $t|_W$. How does $\phi_W(f_P|_W)=t|_W=\phi_W(f_Q|_W)$ follow? I tried drawing the diagram that witnesses the functoriality of $\phi$, but I didn't see how it helps. And probably even a more basic question: why do we need to check this? if we're given some $s_p\in\mathscr F_p'$, it has the form $[(f,U)]$ for some open $U$ that contains $p$ and some $f\in \mathcal F'(U)$. Its preimage under the vertical map is just $f$.
 A: $(\psi_U(\phi_U(s)))_p=0$ holds for every $p$. You need to show $a_p = 0 \> \forall p \implies a = 0.$ That's a good exercise but here is the solution if you need:

 $a_p = 0$ means there exists $V_p \ni p$ such that $a|_{V_p} = 0$. Such $V_p$ cover $U$ and $a|_{V_p} = 0|_{V_p},$ so $a = 0.$

In the proof that $\ker(\psi_U)\subseteq\operatorname{im}(\phi_U)$, you end up with a bunch of $s'_p\in\mathcal{F}'_p, p \in U.$ You want them to come from a single element of $\Gamma(U,\mathcal{F}').$ The idea is that those $s'_p$ are represented by some $f_p \in \Gamma(V_p,\mathcal{F}'),$ and one can use that $\mathcal{F'}$ is a sheaf to glue $f_p$ together to get the desired element. For that, one has to check the compability of $f_p.$ This means showing that $f_p|_W=f_q|_W$ for $W \subset V_p \cap V_q.$ It is enough to show $\phi_W(f_p|_W)=\phi_W(f_q|_W)$ since $\phi_W$ is injective. But here it follows from how $f_p$ are defined. They are defined so that elements $s'_p$ they represent are mapped to $t_p$ by $\phi_p.$
This means that both  $\phi_W(f_p|_W)$ and $\phi_W(f_q|_W)$ locally look like $t$ (so have the same germs) and by the exercise above they must be equal.
