Okay so if I understand you right, you finally want to see why the image presheaf is already a sheaf, so I will give it a try.
(Sorry for giving the morphisms new names.)
If we can show that for $U \subseteq X$ open we have $ker(\psi_U)=im(\phi_U)$, we are done, since then the image presheaf is given by the kernel of a morphism of sheaves, which is indeed a sheaf. So let us try this.
Consider the induced sequence \begin{align*}
0 \longrightarrow \Gamma(U, \mathscr{F}') \overset{\phi_U}{\longrightarrow} \Gamma(U, \mathscr{F}) \overset{\psi_U}{\longrightarrow} \Gamma(U, \mathscr{F}''),
\end{align*} and for each $x \in U$ the induced sequence \begin{align*}
0 \longrightarrow \mathscr{F}'_x \overset{\phi_x}{\longrightarrow} \mathscr{F}_x \overset{\psi_x}{\longrightarrow} \mathscr{F}''_x,
\end{align*} which is exact, since the exactness of a sequence of sheaves is equivalent to the exactness of the induced sequence on stalks for all points. Now putting these together with the natural morphisms into stalks, one gets a commutative diagram (which I unfortunately was not able to type (mea culpa)).
We now come to our claim that $ker(\psi_U)=im(\phi_U)$:
$[\supseteq]:$ Let $s \in \Gamma(U, \mathscr{F}')$ and consider for each $x \in U$ the germ of $\psi_U(\phi_U(s))$ in the stalk $\mathscr{F}''_x$: \begin{align*}(\psi_U(\phi_U(s)))_x = \psi_x(\phi_x(s_x)). \end{align*} But by exactness, $\psi_x(\phi_x(s_x))=0$ for all $x \in U$. Hence $\psi_U(\phi_U(s))=0$, so $im(\phi_U) \subseteq ker(\psi_U)$.
$[\subseteq]:$ Let $t \in ker(\psi_U)$, so $\psi_U(t)=0$. Then for all $x \in U$ we have that $\psi_x(t_x)=(\psi_U(t))_x = 0$, so the germ of $t$ at $x$ is an element in $ker(\psi_x)=im(\phi_x)$ by exactness again. Hence for every $x \in U$ there is a $s'_x \in \mathscr{F}'_x$, say of the form $[(s'_{(x)},V_{(x)})]$ for some open neighborhood $V_{(x)} \subseteq U$ of $x$ and $s'_{(x)} \in \Gamma(V_{(x)}, \mathscr{F'})$, such that $\phi_x(s'_x)=t_x$. Then we have that for $x,y \in U$ \begin{align*}
\phi_{V_{(x)}\cap V_{(y)}}(s'_{(x)} |_{V_{(x)}\cap V_{(y)}}) = t | _{V_{(x)}\cap V_{(y)}} = \phi_{V_{(x)}\cap V_{(y)}}(s'_{(y)} |_{V_{(x)}\cap V_{(y)}}),
\end{align*} so that by the injectivity of $\phi_{V_{(x)}\cap V_{(y)}}$ (which you already proved), we get the required condition \begin{align*}
s'_{(x)} |_{V_{(x)}\cap V_{(y)}} = s'_{(y)} |_{V_{(x)}\cap V_{(y)}}
\end{align*} for the gluing of the $s'_{(x)}$ for $x \in U$. Therefore we have a section $s \in \Gamma(U, \mathscr{F})$ with the property that for all $x \in U$ \begin{align*}
s | _{V_{(x)}} = s'_{(x)}.
\end{align*}
Now we can conclude that for every $x \in U$ \begin{align*}
(\phi_U(s))_x = \phi_x(s_x) = \phi_x(s'_x) = t_x,
\end{align*} since $s_x=s'_x$, which gives $\phi_U(s)=t$ as desired.
Best Answer
Method 1: Use Exercise II.1.4b to relate the image presheaf and its sheafification. Then by exactness of the original sequence, the isomorphism follows on sections of $U$.
Method 2: Proceed just as in proof of Proposition II.1.1.
Let $\varphi: \mathscr{F} \to \mathscr{F''}$ and $\psi : \mathscr{F'} \to \mathscr{F}$. We want to show that the kernel and images are equal after applying $\Gamma (U, \cdot)$. This can be checked on the stalks.
In one direction, we have that
$$(\varphi _U \circ \psi _U (s))_P = \phi _P \circ \psi _ P (s_P) = 0$$
By the sheaf condition, this shows that $\phi _U \circ \psi _ U = 0$.
Conversely, let's suppose $t \in \textrm{ker } \varphi _U$, i.e. $\varphi _U (t) = 0$. Again we know that the stalks is an exact sequence so for each $P \in U$, there is a $s_P$ such that $\psi _P (s_P) = t_P$. Let's represent each $s_P = (V_P , s(P))$ where $s(P) \in \mathscr{F'} (V_P)$. Now, $\psi (s(P))$ and $t \mid _{V_P}$ are elements of $\mathscr{F} (V_P)$ whose stalks at $P$ are the same. Thus, WLOG, assume $\psi (s(P)) = t \mid _{V_P}$ in $\mathscr{F} (V_P)$. $U$ is covered by the $V_P$ and there is a corresponding $s(P)$ on each $V_P$ which, on intersections, are both sent by $\psi$ to the corresponding $t$ on the intersection. Here, we apply injection (exactness at left place, which you showed in your OP) which allows us to glue via sheaf condition to a section $s \in \mathscr{F'} (U)$ such that $s \mid _ {V_P} = s(P)$ for each $P$. Verify that $\psi (s) = t$ and we're done by applying the sheaf property and the construction to $\psi (s) - t$.