I think the étalé space argument from the other answer is nice, that was also the first argument that came to mind while reading the same(?) passage of Serre's book. But when I tried to write it down I found another argument which seemed easier to formalize, so maybe you also find it more convincing.
The claim is in fact that $\mathcal{S}=\bigoplus_{P\in X}\mathcal{S}_{P}$, where the summands on the right are the corresponding skyscraper sheaves.
As you pointed out yourself, there is a dense open subset $U\subseteq X$ over which $\mathcal{S}$ has no sections, so $\mathcal{S}_{P}=0$ for all $P\in U$.
So the direct sum is indeed finite and no sheafification is needed. If $Q\in X$ is a point, then we claim that
$$ \Gamma(V,\mathcal{S})=\bigoplus_{P\in X}\Gamma(V,\mathcal{S}_{P})=\Gamma(V,\mathcal{S}_{Q})=\mathcal{S}_{Q}, $$
where $V$ is an open neighbourhood of $Q$ in $X$ not containing any other point with non-zero stalk.
More precisely, we want to show that the restriction to the stalk $s\mapsto [(V,s)]$ is an isomorphism.
For injectivity, suppose $s\in \Gamma(V,\mathcal{S})$ has $s_{Q}=0$.
Since all the other stalks of $\mathcal{S}$ in $V$ are $0$, we get $s_{P}=0$ for all $P\in V$, hence $s=0$.
For surjectivity, let $t_{Q}=[(W,t)]\in \mathcal{S}_{Q}$ for some $Q\in W\subseteq V$ open and some $t\in \Gamma(W,\mathcal{S})$.
Find a smaller open subset $Q\in W'\subseteq W$ such that $t|_{W'\setminus \{Q\}}=0$ and consider the open cover $V=W'\cup (V\setminus \{Q\})$.
The sections $t|_{W'}\in \Gamma(W',\mathcal{S})$ and $0\in \Gamma(V\setminus\{Q\},\mathcal{S})$ are both $0$ on the intersection, so they glue to a section $\tilde{t}\in \Gamma(V,\mathcal{S})$.
Then $\tilde{t}_{Q}=t_{Q}$, because
$$ [(V,\tilde{t})]=[(W',t|_{W'})]=[(W,t)]. $$
The set of such open sets (as we vary the point $Q$) forms a basis for the topology on $X$.
Since both sheaves have the same sections over a basis of the topology, they have to be equal.
P.S. If this is the part of Serre's Algebraic Groups and Class Fields that I think it is, I believe there is a simpler proof.
With the notation of the book: if $Q\in X$ is a point and $U\subseteq X$ is an open neighbourhood of $Q$ in $X$ such that all points in $U\setminus \{Q\}$ have coefficient $0$ in the divisor $D$, then we can explicitly describe the sections of both $\mathcal{A}$ and of $\bigoplus_{P\in X}\mathcal{A}_{P}$ as equivalence classes of rational functions $f$ on $X$ with $f\sim g$ if and only if $v_{Q}(f-g)+v_{Q}(D)\geqslant 0$.
Hence both sheaves have the same sections over each element of a basis of the topology and thus the two sheaves are the same.
Best Answer
Fist of all, there's a natural isomorphism $$\Gamma (X,\mathcal{F}) = \mathcal{F} (X) \cong \operatorname{Hom}_\mathcal{Ab} (\mathbb{Z}, \mathcal{F} (X)).$$ (And that's where $\mathbb{Z}$ comes from: it's the free abelian group generated by one element.)
Now I claim that $$\operatorname{Hom}_\mathcal{Ab} (\mathbb{Z}, \mathcal{F} (X)) \cong \operatorname{Hom}_{\mathcal{PSh} (X)} (\mathbb{Z}_X, i (\mathcal{F})) \cong \operatorname{Hom}_{\mathcal{Sh} (X)} (\mathbb{Z}_X^\mathbf{a}, \mathcal{F}).$$ Here by $\mathbb{Z}_X$ I denote the constant presheaf on $X$ having $\mathbb{Z}$ as its sections, and by $\mathbb{Z}_X^\mathbf{a}$ its sheafification. The second natural isomorphism is basically the definition of sheafification (as the left adjoint to the inclusion $i\colon \mathcal{Sh} (X) \to \mathcal{PSh} (X)$), and we are interested in the first isomorphism.
We need to check that group homomorphisms $f\colon \mathbb{Z} \to \mathcal{F} (X)$ naturally correspond to presheaf morphisms $\mathbb{Z}_X \to \mathcal{F}$. Such a morphism of presheaves is simply a family of homomorphisms $f_U\colon \mathbb{Z} \to \mathcal{F} (U)$ compatible with the restriction maps for $\mathcal{F}$.
In one direction, having such a family $\{ f_U \}$, we just take the homomorphism $f_X\colon \mathbb{Z} \to \mathcal{F} (X)$.
In the other direction, starting from a group homomorphism $f\colon \mathbb{Z} \to \mathcal{F} (X)$, we may define $\{ f_U\colon \mathbb{Z} \to \mathcal{F} (U) \}$ by taking the compositions of $f$ with the restriction maps $\mathcal{F} (X) \to \mathcal{F} (U)$.
This gives a natural bijection.
Of course, $\mathbb{Z}$ may be replaced with any abelian group: $$\operatorname{Hom}_\mathcal{Ab} (A,\mathcal{F} (X)) \cong \operatorname{Hom}_{\mathcal{PSh} (X)} (A_X, \mathcal{F}) \cong \operatorname{Hom}_{\mathcal{Sh} (X)} (A_X^\mathbf{a}, \mathcal{F}).$$ So we just saw that the global section functor is right adjoint to the constant (pre)sheaf functor.