The trick to translating a set into a universal property is to note that "the set of all $P$" can be pieced apart:
- $P$ is where you get the shape of your diagram
- "set of all" is the universal part of the universal property
Just to illustrate, for $X$ and $Y$ sets, $X\times Y$ is the "set of all pairs $(x,y)$ for $x\in X$ and $y\in Y$", the pairs part being the diagram $\{X\quad Y\}$, then "set of all" is taking the limit of this diagram.
In this case, we're trying to take the set of all commutative squares with $f$ and $g$ parallel.
The interpretation of the result you state about a limit being an equaliser of products tells you that an arbitrary limit can be constructed (in $\mathbf{Set}$) by collecting all the ingredients (i.e., form those products) and then subject them to constraints (i.e., form the equaliser).
For a pullback of $A\xrightarrow pB\xleftarrow qC$, the equaliser diagram is
$$
A\times_BC \to A\times B\times C \rightrightarrows B\times B
$$
where the maps $A\times B\times C\to B\times B$ are given by $A\times B\times C\xrightarrow\pi A\times B\xrightarrow{p\times B} B\times B$ (where $\pi$ is a projection) and similarly for $q$.
You might notice this is a bit redundant, and we can really take the pullback to be the equaliser
$$
A\times_BC \to A\times C\rightrightarrows B
$$
of the maps $A\times C\to A\to B$ and $A\times C\to C\to B$.
In any case, this allows us to interpret the pullback $A\times_BC$ as "the set of all pairs $(a,c)$ for which $p(a)=q(c)$".
Now, what is a commutative square involving $f$ and $g$? Well, it's a pair of morphisms $u:a\to c$ and $v:b\to d$ such that
$\require{AMScd}$
\begin{CD}
a @>u>> c \\
@VfVV @VVgV \\
b @>>v> d
\end{CD}
commutes; that is, $g\circ u=v\circ f$. So, $\operatorname{Sq}(f,g)$ is "the set of all pairs $(u,v)$ such that $g\circ u=v\circ f$," which is exactly what a pullback describes!
To finish the job, we just need to specify the sets. $u$ comes from $\def\Hom{\operatorname{Hom}}\Hom(a,c)$ and $v$ from $\Hom(b,d)$, while the two composites lie in $\Hom(a,d)$, so we get the pullback square
\begin{CD}
\operatorname{Sq}(f,g) @>>> \Hom(a,c) \\
@VVV @VVg\circ(-)V \\
\Hom(b,d) @>>(-)\circ f> \Hom(a,d)
\end{CD}
This is a duplicate question, but I cannot find the duplicate right now.
It has been already explained why locally small categories are important (tons of examples, and the Yoneda Lemma). Examples of categories which are not locally small have been given at SE/219539 and MO/3278. Here is an attempt to answer the set-theoretic concerns.
First of all, Grothendieck universes are very useful to manage size-isszes in category theory. For example, locally small categories can be considered to be small in this way. In detail:
Let $U$ be a Grothendieck universe. A $U$-small (or just: small) category is a category such that both morphisms and objects form sets in $U$, also called $U$-small sets.
Many categories in practice are not small, but they are locally small. But in order to formalize this, it is best to assume another Grothendieck universe $U \in V$. Then we define* a locally small category to be a $V$-small category such that all hom-sets are $U$-small. Notice that such a category is still small, just with respect to a larger Grothendieck universe. This is very useful, since you can apply all of the known constructions of small categories also to all categories, when you keep track that you are working with $V$-small categories. In particular, we can consider functor categories for $V$-small categories. So from this perspective, there are no "large categories".
If $\mathbf{Set}$ denotes the category of $U$-small sets, for every locally small category $\mathcal{C}$ we have the Yoneda embedding $\mathcal{C} \to \mathrm{Hom}(\mathcal{C}^{\mathrm{op}},\mathbf{Set})$, $A \mapsto \mathrm{Hom}(-,A)$. Notice that both sides here are $V$-small. The Yoneda embedding is very important, since it reduces many claims from an arbitrary locally small category to the category of sets.
*Unfortunately, this is not the only definition in the literature. See MO/3409.
Best Answer
Let rewrite the indicies in the diagram you gave : $$ \prod_{k \in Ob(J)}\mathbf{C}(Fk,Gk)\rightrightarrows\prod_{f:i\to j}\mathbf{C}(Fi,Gj). $$
By universal property, a map into a product is fully described by the maps to it's components.
Let $f:i\to j$ a morphism in $\mathbf C$.
The $f$-component of the first map is given by the composition : $$ \prod_{k} \mathbf{C}(Fk,Gk) \to \mathbf{C}(Fi,Gi)\to \mathbf{C}(Fi,Gj) $$ where the first is the $i$ projection, and the second is composing with $G(f)$, namely the map $G(f)_*$.
The $f$-component of the second map is given by the composition : $$ \prod_{k} \mathbf{C}(Fk,Gk) \to \mathbf{C}(Fj,Gj)\to \mathbf{C}(Fi,Gj) $$ where the first is the $j$ projection, and the second is precomposing with $F(f)$, namely the map $F(f)^*$.
So we get the two desired maps of the diagram. Then an element that equalizes those two maps should be an element $(\alpha_k)_{k \in Ob(J)}$ of $\prod_{k} \mathbf{C}(Fk,Gk)$ such that for any $f : i \to j$, $F(f)^* (\alpha_j) = G(f)_* (\alpha_i)$, i.e. $$ \alpha_j \circ F(f) = G(f) \circ \alpha_i $$ which is the equation defining natural transformations. So, by definition, the equalizer of those two maps is $Nat(F,G)$.