I've already proven that $(\ker \varphi)_p \subset \ker \varphi_p$ using a commutative diagram and the definition $F_p = \lim\limits_{\longrightarrow \\ U \ni p} F(U) = \bigsqcup\limits_{U \ni p} F(U)/ \sim$ where $[x] = [y]$ in the equivalence relation $\iff \rho_{UW}(x) = \rho_{VW}(y)$ for some $W \subset U \cap V$ open sets in the topology $X$. I got that from Wikipedia article for direct limit. Where $\rho_{UV}$ are the restriction maps of the sheaf $F$.
So I have a morphism of sheaves $\varphi : F \to G$, which induces $\varphi_p : F_p \to G_p$. Intuitively it seems like if we assume $[x] \in \ker \varphi_p$ it's going to take a "subtraction trick" to show that $y \sim x$ for some $y \in \ker \varphi(U)$. I'm just not sure where to deploy this trick.
The sheaves are abelian group-valued.
Any ideas?
The red and the blue represent two different diagram chases. First we assume (blue) $[x] \in \ker \varphi_p$. That's as far as I've gotten and shows that I did some work. :)
Diagram drawn using: https://www.mathcha.io/
$j$ is the inclusion map I proved in the converse direction on paper.
Suppose $x \in F_p$ is in $\ker(\varphi_p)$. Then there exist a neighborhood $U$ of $p$ and $\bar x \in F(U)$ such that $(\bar x)_p = x$. Furthermore, since $(\bar x)_p \in \ker(\varphi_p)$, we have that $[\varphi_U(\bar x)]_p = 0 \in G_p$. Therefore, there exists some neighborhood $V \subseteq U$ of $p$ such that $[\varphi_U(\bar x)] |_V = 0$. But then $[\varphi_U(\bar x)]|_V = \varphi_V(\bar x |_V)$. So, $\bar x |_V \in [\ker(\varphi)](V)$, and $[\bar x |_V]_p = x$.