You ask: "Why must one sheafify the presheaf $P(\mathcal{F}/\mathcal{G})$ then?". The answer is: "Because it is not a sheaf!"  Here is an example.

Let $X=\mathbb P^1_k, \mathcal F = \mathcal O, \mathcal G=\mathcal O (-2.P)$, where $P$ denotes the origin $O=(0:1)$ .  Define $\infty=(1:0)$ and let $z$ be the coordinate on $\mathbb P^1\setminus \infty$ .

 Consider the covering of X by the open subsets $U_0=X\setminus \infty$ and $U_\infty =X\setminus O$. Let me denote the presheaf $P(\mathcal{F}/\mathcal{G})$ just by $P$.

Then $class(z)\in P(U_0)$ and $class(0)\in P(U_\infty)$ are sections of $P$ over the two open sets $U_0$ and $U_\infty$ of our covering which coincide on  their intersection $U_{0\infty}$ for the excellent reason that $P(U_{0\infty})=0$ ! [Actually $P(U)=0$ for any open subset $U\subset X$ not containing $O$] 

But these compatible sections cannot be glued to a global section of $P$ on $X$. Indeed a section of $P$ on $X$ is just a constant $c\in k$ , since $\mathcal O(X)=k$ and $\mathcal O(-2.P)(X)=0$.
But that constant $c$ cannot be the glued global section , because its restriction to  $P(U_0)$ is $ class(c) $ and in $P(U_0)$ we have $class(c) \neq class(z)$ since $z-c$ doesn't vanish with order  two at $P$.