I was trying to prove that $\varinjlim_{U \ni x} A(U) \cong A_{\mathfrak{p}}$ where $U$ are basic open sets containing $x = \mathfrak{p} \in \operatorname{Spec} A$, and $A(U) = A_f$ where $U = X_f$. $A$ is a commutative ring.
I know that the restriction maps from $A(X_f)$ to $A(X_g)$ are given by $a/f^k \mapsto au^k/g^{nk}$ if $g^n = uf$ for some $u \in A$, some $n >0$. But in this case I need to know explicitly how the maps from $A_f$ to $A_{\mathfrak{p}}$ are given. I have constructed maps $\alpha_f: A_f \to A_{\mathfrak{p}}$ by factoring the canonical map $A \to A_{\mathfrak{p}}$ through the canonical map $A \to A_f$, and showed that they commute with the direct system, thus we have some unique commuting homomorphism $\alpha : \varinjlim_{U\ni x}A(U) \to A_{\mathfrak{p}}$.
To show surjectivity of this homomorphism, I need to know explicitly the map $\alpha_f$, which I presumed to be $a/f^k \mapsto a/f^k$, but I am seriously doubting this. This map is well-defined and is a homomorphism (in fact it's just an inclusion) but is it safe to conclude that our desired $\alpha_f$ is this map, using uniqueness property of localization? I have a vague feeling that I made a very serious mistake here. I would appreciate if anyone points out.
That's the correct definition of $\alpha_f$, yes. Really $A_f\to A_{\mathfrak{p}}$ is just localisation of $A_f$ at $\mathfrak{p}A_f$, though it is worth noting the homomorphisms in question need not be injective (this requires $A\setminus\mathfrak{p}$ to have no zero divisors, which might not be true) and thus they don't deserve to be called 'inclusions'.
On one hand you can use the universal property of localisation to conclude that some factoring $A_f\to A_{\mathfrak{p}}$ must exist; on the other hand, you can use the proof that the universal property holds ! or directly inspect that $\alpha_f$ so defined works, to see that it works. Like, you don't need to conclude $\alpha_f$ is what it is, per se, you can just give it and see it works. We must send $1/f$ in $A_f$ to the inverse of $f$ in $A_{\mathfrak{p}}$. Oh, but we just call that $1/f$ as well, so the map really is just (induced by) $a/f\to a/f$.
To see $A_{\mathfrak{p}}\cong\varinjlim_{f\notin\mathfrak{p}}A_f\cong\varinjlim_{U\ni x}\mathcal{O}_{\mathrm{Spec}(A)}(U)$ you just have to check they all have the correct universal properties. The second isomorphism is straightforward as each $U\ni x$ contains some $D(f)$ for some $f$ (we get a final, or cofinal, whatever, subcategory, which preserves colimits) and the first isomorphism just (morally) comes from the fact that a map $A\to B$ inverts every element not in $\mathfrak{p}$ iff. ... well, iff. it inverts every element not in $\mathfrak{p}$ and therefore factors through every $A_f$ (in compatible ways)! It is also important to know these are all isomorphisms under $A$, but this is obvious if you keep track of what the universal properties really mean.
As requested, let's touch on the categorical perspective a little more carefully. For the final subcategory thing: if $\mathcal{O}(X)_x$ is the poset of nonempty open subsets of $X$, containing $x$, ordered by containment, then the full subcategory of $\mathcal{O}(X)_x$ on some basis $\mathcal{B}$ of $X$, $\mathcal{O}_{\mathcal{B}}(X)_x$, is a final subcategory or equivalently $\mathcal{O}_{\mathcal{B}}(X)_x\to\mathcal{O}(X)_x$ is a (fully faithful and) final functor. A functor $F:\mathscr{C}\to\mathfrak{D}$ is said to be final if, for all diagrams $H:\mathfrak{D}\to\mathscr{A}$ admitting a colimit, the natural comparison map $\varinjlim_{\mathscr{C}}HF\to\varinjlim_{\mathfrak{D}}H$ is an isomorphism. When you have a nice directed poset category like $\mathcal{O}(X)_x$ checking finality is very easy. Or cofinality, whatever, the usage seems to a vary a bit. Essentially, if you have a cocone $H\implies\Delta A$ for some object $A\in\mathscr{A}$ then its components are determined by the components on the basis $\mathcal{B}$; if $x\in U\subseteq X$ is any open, there is surely a basis element $V\in\mathcal{B}$ with $x\in V\subseteq U$ and the map $H(U)\to A$ is necessarily equal to $H(U)\to H(V)\to A$ and is thus determined by $H(V)\to A$; using this observation it's not too hard to argue we have a final subcategory.
For the stalk calculation: firstly, as $\mathfrak{p}\subset A$ is proper, there surely exists at least one $f_0\in A\setminus\mathfrak{p}$. Let's keep track of this guy; the choice doesn't matter but it's going to be useful to have a reference point. For any $f\notin\mathfrak{p}$ let $\lambda_f$ denote the colimit cocone $A_f\to\varinjlim_{g\notin\mathfrak{p}}A_g$, where of course the colimit is computed in the category $\mathsf{CRing}$ of commutative unital rings. The canonical map $A\to\varinjlim_{f\notin\mathfrak{p}}A_\mathfrak{p}$ is $A\to A_{f_0}\overset{\lambda_{f_0}}{\longrightarrow}\varinjlim_{f\notin\mathfrak{p}}A_f$. Since for any other $f_1$, the following commutes: $$\require{AMScd}\begin{CD}A@>>>A_{f_0}\\@VVV@VVV\\A_{f_1}@>>>A_{f_0f_1}\end{CD}$$(it is important now and with all manipulations of stalks to note and use the fact that the diagram of which the colimit is taken is directed)And the maps $\lambda_{f_0},\lambda_{f_1}$ factor through $\lambda_{f_0f_1}$ by definition of cocone, the composite $A\to\varinjlim_{f\notin\mathfrak{p}}A_f$ is independent of the choice of $f_0$.
Say $\psi:A\to B$ is a map of $\mathsf{CRing}$ such that $\psi(f)$ is a unit for every $f\notin\mathfrak{p}$. Then for every $f$, we know $\psi$ factors as some composite $A\to A_f\overset{\varphi_f}{\longrightarrow}B$. Moreover if $g\in\sqrt{(f)}$ (iff. $D(g)\subseteq D(f)$) then we have $\varphi_f$ equal to $A_f\to A_g\overset{\varphi_g}{\longrightarrow}B$. Why? Because both, when postcomposed with $A\to A_f$, yield the same thing ($\psi$) - this is because $A\to A_f\to A_g$ equals $A\to A_g$ - and the uniqueness part of the universal property gives me what I want. This compatibility observation tells me that the $(\varphi_f)_{f\notin\mathfrak{p}}$ form a cocone under the relevant diagram hence factor through the $\lambda_f$ and some $\Gamma:\varinjlim_{f\notin\mathfrak{p}}A_f\to B$.
By construction of $\Gamma$, observe: $$\begin{align}A\to\varinjlim_{f\notin\mathfrak{p}}A_f\overset{\Gamma}{\longrightarrow}B&=A\to A_{f_0}\overset{\lambda_{f_0}}{\longrightarrow}\varinjlim_{f\notin\mathfrak{p}}A_f\overset{\Gamma}{\longrightarrow}B\\&=A\to A_{f_0}\overset{\varphi_{f_0}}{\longrightarrow}B\\&=\psi\end{align}$$
Alright, so $\psi$ indeed factors through $A\to\varinjlim_{f\notin\mathfrak{p}}A_f$ by some $\Gamma$. Is $\Gamma$ unique? Well, if $\Gamma'$ has this property then $\Gamma'\circ\lambda_f$ and $\varphi_f$ both, when postcomposed with $A\to A_f$, yield $\psi$, so by the uniqueness in the universal property of localisation we know $\Gamma'\circ\lambda_f=\varphi_f$; furthermore, that this is true for every $f\notin\mathfrak{p}$ gives, by the uniqueness in the universal property of colimit, that $\Gamma'=\Gamma$. So yes, it is unique.
An important but obvious point: we also need to note that the map $A\to\varinjlim_{f\notin\mathfrak{p}}A_f$ does in fact map every single $g\notin\mathfrak{p}$ to a unit. This is true because this map equals $A\to A_g\overset{\lambda_g}{\longrightarrow}\varinjlim_{f\notin\mathfrak{p}}A_f$ for any such $g$ (we checked that already) and $g$ is mapped to a unit by the first component, and $\lambda_g$ certainly preserves units (any unital ring map does) so overall $g$ is indeed mapped to a unit. Now all boxes of the universal property of localisation are ticked for $A\setminus\mathfrak{p}$, so we conclude $A\to\varinjlim_{f\notin\mathfrak{p}}A_f$ is a localisation of $A$ at $A\setminus\mathfrak{p}$ thus is a model for, and is uniquely isomorphic under $A$ to, $A_\mathfrak{p}$ as more standardly constructed.
We probably could save some effort here by computing this as a colimit over some slice category, because it is true and essential that all maps involved are maps under $A$. I haven't thought about this though.