This comes from exercise II.1.2(a) from Hartshorne. I am stuck with trying to show the inclusion $ker(\phi_{p}) \subseteq ker(\phi)_{p}$ where we have the morphism of sheaves given by $$\phi: \mathcal{F} \to \mathcal{G}$$ I understand that by definition the stalk is given by equivalence classes of pairs $<U,s>$, where $U \ni p$ is an open neighborhood of the topological space $X$, and $s$ is a section in $\mathcal{F}(U)$.
How should I prove the above inclusion? Does it not easily follow that the existence of a pair $<U,s> \in \ker(\phi_{p}) \subseteq \mathcal{F}_{p}$ imply that we can find an open neighborhood $U \ni x$ and the section $s \in \mathcal{F}(U)$ such that it vanishes on $\phi(U)$?
In particular, I have been looking at the solution given here: http://www.math.utah.edu/~zwick/Classes/Hartshorne/Section2_1.pdf
I don't understand the statement $s|p = s_{p}$. Isn't $s_{p} \in \mathcal{F}_{p}$ an ordered pair while $s|p$ is the restriction of the section to a single point? And I also don't understand why that implies the second statement and the existence of the equation on an open set when the equality only holds at a single point $p$.
Edit: I guess the main thing that I'm not solid on is what exactly the notion of the germ $s_{p}$, i.e. the image of the section $s$ in $F_{p}$. Is it simply fixing the above pair $<U,s>$ to a single section $s_{o}$, whilst keeping the same equivalence relations, which is just the notation $s|p$, along with the associated open set?