I am doing the Exercise 4.7.20 in Bas Edixhoven's lecture notes of Algebraic geometry. Below is the statement of the exercise, and I am attempting part (iii).
(Recall for the notation: $\mathcal{O}_X(U)$ is the $k$-subalgebra of the functions $s: U \to k$ which are regular locally on $U \cap V$ where $V$ is an open subset of $\mathbb{A}^n$. $\mathcal{O}_X^{\rm p}(U)$ is the $k$-subalgebra of the functions $s: U \to k$ which are regular globally on $U$.)
My attempt is trying to prove the universal property directly using the map of stalks in the part (ii). Firstly let $\mathcal{G}$ be a sheaf and $G: \mathcal{O}_X^{\rm p} \to \mathcal{G}$ be a morphism of (pre)sheaves, then for all inclusion $U \subseteq V$ the following diagram commutes:
$\require{AMScd}$ \begin{CD} \mathcal{O}_X(V) @<{i_V}<< \mathcal{O}_X^{\rm p}(V) @>{G_V}>> \mathcal{G}(V)\\ @V{r_{V,U}}VV @V{r_{V,U}}VV @VV{r_{V,U}}V\\ \mathcal{O}_X(U) @<<{i_U}< \mathcal{O}_X^{\rm p}(U) @>>{G_U}> \mathcal{G}(U) \end{CD}
From part (ii) I also have a bijection $\Psi: \mathcal{O}_{X,x}^{\rm p} \xrightarrow{\simeq} \mathcal{O}_{X,x}$ defined by $[(U,s)]_{\rm p} \mapsto [(U,i_U(s))]=[(U,s)]$. Here the equivalence class symbol with and without p on the feet to distinguish elements on different stalks.
My idea is for every open subset $U$ of $X$, I map $s \in \mathcal{O}_X(U)$ to $[(U,s)] \in \mathcal{O}_{X,x}$ and then to $\Psi^{-1}([(U,s)]) \in \mathcal{O}_{X,x}^{\rm p}$. But I don't know how to go on to $\mathcal{G}(U)$, since in part (ii) when I showed that $\Psi$ is surjective, I chose the pre-image of $[(U,s)]$ as $[(U \cap W,s_{\mid U \cap W})]_{\rm p}$, where $W$ is an open neighborhood of $\mathbb{A}^n$ in the definition of $s \in \mathcal{O}_X(U)$.
At the moment I prefer a direct proof of universal property, but any other approaches are still fine.
Any help will be really appreciated.
As per the OPs request, I'm posting a proof of the universal property, i.e., not assuming the existence of a sheafification beforehand. We will need some preliminary results to structure the proof and to not repeat the same argument over and over again, even though they might seem obvious once one has internalised the power of stalks on sheaves. (Added in proof: That turned out to be more lengthy than I had anticipated, but I don't see a way to shorten it without assuming the yoga of stalks and their relationship with the sheafification, which would be cheating since it renders the claim obvious altogether, defeating the whole point of it.)
Throughout, let $X$ be a topological space.
Proposition: Let $i\colon A\to B$ be a morphism of pre-sheaves on $X$ where $B$ is a sheaf. Then following are equivalent:
Recall that a pre-sheaf $A$ on $X$ is called separated if for every $U\subset X$ open, a pair of local sections $r,s\in A(U)$ agrees if (and only if) they agree on an open cover, i.e., if there is an open cover $\bigcup_{i\in I}U_i=U$ such that $r|_{U_i}=s|_{U_i}$ for all $i\in I$. Clearly, a sheaf is a separated pre-sheaf.
Lemma 1: Let $A$ be a pre-sheaf on $X$. Then $A$ is a separated pre-sheaf if and only if for every open set $U\subset X$ and every pair of local sections $r,s\in A(U)$ such that $r_x=s_x$ for all $x\in U$, we have $r=s$.
Proof. Since we will only need one implication, we will skip the (easy) proof of the other. Let $A$ be a separated pre-sheaf and suppose $r,s\in A(U)$ agree on all stalks $r_x=s_x\in A_x$, $x\in U$. Then, by the definition of the stalk, around each $x\in X$ there exists an open neighbourhood $U_x$ such that $s|_{U_x}=r|_{U_x}$. But the $U_x$ cover $U$ and so $r=s$ since $A$ is separated, as claimed.
Corollary 1: Morphisms into separated pre-sheaves are uniquely determined on stalks, i.e., if $\varphi,\psi\colon A\to B$ is a pair of pre-sheaf morphisms and $B$ is separated, then $\varphi=\psi$ if and only if $\varphi_x=\psi_x$ for all $x\in X$.
Proof: We have to show that $\varphi(s)=\psi(s)$ for all $U\subset X$ open and all $s\in A(U)$. But by definition of the map induced on stalks and by assumption, we see that for all $x\in U$, $\varphi(s)_x=\varphi_x(s_x)=\psi_x(s_x)=\psi(s)_x$. Since $B$ is separated, Lemma 1 gives the claimed equality $\varphi(s)=\psi(s)$.
Corollary 2: Let $A, B$ be a pre-sheaves on $X$ where $B$ is separated. Then a collection of maps on stalks $\varphi_x\colon A_x\to B_x$, $x\in X$, is induced from a (necessarily unique) morphism $\varphi\colon A\to B$ if and only if there exists a collection of maps $\{\varphi'_U\colon A(U)\to B(U)\}_{U\subset X\text{ open}}$ such that $\varphi'_U(s)_x=\varphi_x(s_x)$ for all $x\in U\subset X$ and every local section $s\in A(U)$. Moreover, in that case, $\varphi_{U}=\varphi'_U$ for all open subsets $U\subset X$.
Proof: The "moreover"-part at the end (and consequently the uniqueness-claim) follows from Corollary 1 and the "only if"-part is trivial. Conversely, suppose we are given maps $\varphi_x\colon A_x\to B_x$, $x\in X$, as well as $\varphi'_U\colon A(U)\to B(U)$, $U\subset X$ open, satisfying the stated compatibility condition. Then, for each pair of open subsets $V\subset U\subset X$ and all $s\in A(U)$, we conclude from Lemma 1 that $\varphi'_U(s)|_V=\varphi'_V(s|_V)$, for $$(\varphi'_U(s)|_V)_x=\varphi'_U(s)_x=\varphi_x(s_x)=\varphi_x((s|_V)_x)=\varphi'_V(s|_V)_x$$ for each $x\in V$. Therefore, the $\varphi'_{\bullet}$ do define a morphism $\varphi'\colon A\to B$. Clearly, $\varphi'_x=\varphi_x$ for all $x\in X$ and so the claim follows.
Lemma 2: Let $\varphi\colon A\to B$ be a morphism of pre-sheaves on $X$ such that $\varphi_x\colon A_x\to B_x$ is surjective for all $x\in X$. Then for each local section $s\in B(U)$ over some open set there exists an open cover $U=\bigcup_{i\in I}U_i$ and a collection of local sections $r_i\in A(U_i)$ such that $\varphi(r_i)=s|_{U_i}$ for all $i\in I$.
Proof. In the setup of the claim, for every $x\in U$, there exists an $r_x\in A_x$ such that $\varphi_x(r_x)=s_x$. Choose a representative of $r_x$ on an open cover $U_x$ of $x$, say $r_{U_x}\in A(U_x)$. Then $\varphi(r_{U_x})_x=\varphi_x((r_{U_x})_x)=\varphi_x(r_x)=s_x$; thus, after shrinking $U_x$ to a smaller open neighbourhood of $x$, if necessary, we can arrange that $\varphi(r_{U_x})=s|_{U_x}$. Since the $U_x$ cover $U$, this concludes the proof.
Proof of the Proposition: Again, since the OP only requires the proof of $2.\Rightarrow 1.$, we will skip the (not entirely obvious) proof of the converse implication. Let $i\colon A\to B$ be a morphism of pre-sheaves on $X$ where $B$ is a sheaf and which induces isomorphisms on all stalks $i_x\colon A_x\to B_x$, $x\in X$. Moreover, let $\varphi\colon A\to C$ be a morphism to a sheaf $C$ on $X$. We have to show that there exists a unique morphism $\psi\colon B\to C$ such that $\varphi=\psi\circ i$.
Uniqueness follows from Corollary 1, for any morphism $\psi\colon B\to C$ such that $\varphi=\psi\circ i$ necessarily satisfies $\psi_x=\varphi_x\circ i_x^{-1}$ for all $x\in X$.
Let $U\subset X$ be an arbitrary open subset and let $s\in B(U)$ be an arbitrary local section. By Lemma 2, there exists an open cover $U=\bigcup_{j\in J}U_j$ and a collection of sections $r_j\in A(U_j)$ such that $i(r_j)=s|_{U_j}$ for all $j\in J$. Note that the value $$\varphi(r_j)_x=\varphi_x((r_j)_x)=\varphi_x(i_x^{-1}\circ i_x((r_j)_x))=\varphi_x(i_x^{-1} s_x)$$ in $C_x$ is independent of $j\in I$. In particular, the local sections $\varphi(r_j)$, $j\in J$, agree on pairwise intersections of the $U_j$ by Lemma 1. Thus, since $C$ is a sheaf, the collection $\{\varphi(r_j)\}_{j\in J}$ uniquely glues to a section that shall be denoted by $\psi_U(s)\in C(U)$. Since $U$ and $s\in B(U)$ were arbitrary, we have constructed a collection of maps $\{\psi_U\}_{U\subset X\text{ open}}$ satisfying $\psi_U(s)_x=\varphi_x\circ i_x^{-1} s_x$ for all $x\in U$ and $s\in B(U)$. Therefore, by Corollary 2, the $\psi_U$ assemble into a morphism $\psi\colon B\to C$ satisfying $\psi_x=\varphi_x\circ i^{-1}_x$ or, equivalently, $(\psi\circ i)_x=\varphi_x$ for all $x\in X$. By Corollary 1, this implies $\varphi=\psi\circ i$ and the proof is, finally, complete.