Here are my reflections on the specific sheafification construction in Hartshorne. First, the paragraph in question:
We construct the sheaf $\overline{\mathcal F}$ as follows. For any open set $U$, let $\overline{\mathcal F}(U)$ be the set of functions $s$ to the union $\bigcup_{P\in U}\mathcal F_P$ of the stalks of $\mathcal F$ over points of $U$, such that
- for each $P\in U$, $s(P)\in \mathcal F_P$, and
- for each $P\in U$, there is a neighbourhood $V$ of $P$, contained in $U$, and an element $t\in \mathcal F(U)$, such that for all $Q\in V$, the germ $t_Q$ of $t$ at $Q$ is equal to $s(Q)$.
So, what is going on here? I like to think about the (ring-)sheaf $\mathcal F$ of analytical functions on $\Bbb C$ with the standard topology, because it's a very nice sheaf. Let's sheafify it! (Even though it's already a sheaf; we shall take a sub-presheaf afterwards and go through the same process.)
In order to sheafify, we need to know what the stalks are. In this case, the stalk $\mathcal F_P$ over a point $P$ is the ring of power series around $P$ with positive radius of convergence. For any element $s\in \mathcal F(U)$, the germ $s_P\in \mathcal F_P$ is the power series of $s$ expanded at $P$.
Let's first see what we get by only imposing 1. above, and not 2. This fives us a sheaf $\mathcal F'$ that consists of functions $s:U\to \bigcup_{P\in U} \mathcal F_P$ with only the restriction that $s(P)\in \mathcal F_P$. Where does that leave us? It means that a section of $\mathcal F'(U)$ consists of picking one power series at each point in $U$, with only the restriction that each power series should have positive radius of convergence. This is, in other words, a pretty large sheaf (it has a name: the Godement sheaf or Godement resolution of $\mathcal F$).
So, let's take this large sheaf $\mathcal F'$, and impose restriction 2, which gives us $\overline{\mathcal F}$. What does this restriction really say? It imposes some coherence restrictions on what power series we allow in a section. It says that given a section $s\in \overline{\mathcal F}(U)$, for each point $P$ there is a neighbourhood $V\subseteq U$ of $P$ and an analytic function $t$ on $V$ (this function is from $\mathcal F$, mind you!) such that for each point in $V$, the power series that $s$ defines at that point is the power series of $t$ at that point. In other words, in $\overline{\mathcal F}(U)$ we place the restriction that the power series given at each point by a section should all come from the same analytic function.
Now, what happens if we take a sub-presheaf and work with that instead? This will show you how sections are added by sheafification, although sadly it won't show you how superfluous sections are killed.
Take the presheaf $\mathcal F$ of analytic functions on $\Bbb C$ with an antiderivative, and let $U = \Bbb C-\{0\}$. Thus, for instance, we have $\frac1z \notin \mathcal F(U)$. However, $\frac1z$ does have an antiderivative on $\Bbb C-(-\infty, 0]$ and on $\Bbb C-[0, \infty)$, which means that a section corresponding to $\frac1z$ could be made locally. This shows that $\mathcal F$ is not a sheaf (and incidentally, is the standard example of the fact that image presheafs of a map of sheafs isn't always a sheaf; $\mathcal F$ is the image presheaf of the differentiation map on the sheaf of analytical functions).
So, how does this section get added to our $\overline{\mathcal F}(U)$? The simple answer is that the function $\frac1z$, while it doesn't have an antiderivative on $U$, it does have a power series with positive radius of convergence on each point of $U$, and for each such point $P$, there is a neighbourhood $V\ni P$ such that $\frac1z$ has an antiderivative on $V$. That means that $\frac1z\in \mathcal F(V)$, and this plays the role of our $t$ above.
As for how locally zero sections of presheafs get killed, they die once you pass to stalks. If $s\in \mathcal F(U)$ is non-zero, but locally zero, then for each point $P\in U$, the germ $s_P\in \mathcal F_P$ is zero. And since the image of $s$ over the canonical map $\mathcal F(U)\to \overline{\mathcal F}(U)$ is the function that to each point $P\in U$ assigns the germ $s_p\in \mathcal F_P$, this becomes the zero function.