Proving Proposition I.5.1 of Mac Lane and Moerdijk. This is Exercise I.11 of Mac Lane and Moerdijk's, "Sheaves in Geometry and Logic [. . .]".
The Details:
Adapted from p. 25, ibid. . . .

Definition: Let $\mathbf{C}$ be a category. Then $\hat{\mathbf{C}}=\mathbf{Sets}^{\mathbf{C}^{{\rm op}}}$ is the category of presheaves of $\mathbf{C}$.

Adapted from p. 26, ibid. . . .

Definition 2: Presheaves which are, up to isomorphism, of the form $\mathbf{y}(C)$, where
$$\begin{align}
\mathbf{y}: \mathbf{C}& \to \mathbf{Sets}^{\mathbf{C}^{{\rm op}}}\\
C&\mapsto {\rm Hom}_{\mathbf{C}}(-, C)
\end{align}$$
is the Yoneda embedding, are called representable presheaves or representable functors.

From p. 41 ibid. . . .

Proposition I.5.1. In a functor category $\mathbf{Sets}^{\mathbf{C}^{{\rm op}}}$, any object $P$ is the colimit of a diagram of representable objects, in a canonical way.

From p. 149 onward of Mac Lane's, "Categories for the Working Mathematician," we have . . .

Definition 3: A fork in a category $\mathbf{C}$ is a diagram
$$a \overset{\partial_0}{\underset{\partial_1}{\rightrightarrows}}b\stackrel{e}{\rightarrow}c$$
such that $e\partial_0=e\partial_1$. [. . .] A split fork is a fork with two more arrows
$$a\stackrel{t}{\leftarrow}b\stackrel{s}{\leftarrow}c,$$
where $e\partial_0=e\partial_1$, $es=1_c$, $\partial_0t=1_b$, and $\partial_1t=se$. [. . .] By a split coequaliser of $\partial_0$ and $\partial_1$, we mean the arrow $e$ in a split fork.

The Question:

Prove Proposition I.5.1, that every functor $P$ is representable, by constructing for each $P:\mathbf{C}^{{\rm op}}\to\mathbf{Sets}$, a coequaliser 
$$\coprod_{\begin{array}{c} C'\stackrel{u}{\to}C \\ p\in P(C)\end{array}}\mathbf{y}(C')\overset{\theta}{\underset{\tau}{\rightrightarrows}}\coprod_{\begin{array}{c} C\in \mathbf{C} \\ p\in P(C)\end{array}}\mathbf{y}(C)\stackrel{\epsilon}{\to} P, $$
where $\coprod$ denotes the coproduct and for each object $B$ the maps are defined for each $v: B\to C$ or $C'$ as follows 
$$\begin{align}
\epsilon_B(C, p; v)&=P(v)p,\\
\theta_B(u, p; v)&=(C, p; uv), \\
\tau_B(u,p;v)&=(C', pu; v).
\end{align}$$
[Hint: For each $B$, this gives a split coequaliser.]

Thoughts:
(Phew! That took ages to type up!)
This is all a big mess of symbols to me at the moment. I was hoping that, by asking the question here, I'd get some idea of what's going on as I relay the information; there's very little luck so far though!
I get that each "$B$-component" of $\epsilon, \theta$, and $\tau$ needs three arguments, as described.
Colimits are fairly easy to understand.
On p. 42 of "Sheaves [. . .]", there's 

Corollary I.5.3 (= Proposition I.5.1). Every presheaf is a colimit of representable presheaves.

There's a proof of this Corollary. However, I am not sure how it translates to Exercise I.11.
Please help :)
 A: The way you should understand this problem is the following : 
To illustrate everything I am going to say, I will take the simplest example I know of, by considering the category $\textbf{G}$, which has two objects that I will denote $0$ and $1$, and two morphisms $0\to 1$, and that's it. A presheaf $X$ over $\textbf{G}$ consists of two sets $X_0$ and $X_1$ together with two applications $X_1 \to X_0$. In other words, these presheaves are (multi)graphs, with $X_0$ the set of vertices, $X_1$ the set of arrows, and the two arrows are the source and the target.
The presheaves in the category $\hat{\textbf{C}}$ are built by glueing together some "building blocks", and the category $\textbf{C}$ gives you a description of these building blocks. In my example of graphs, the category $\textbf{G}$ describes two types of building blocks, corresponding to vertices and arrows, along with two relations between them corresponding to the source and target. 
Given a presheaf $P$, how to know how to build it from the building block? If you have an object $c$ of the category $\textbf{C}$, the set of building blocks of type $c$ in $P$ is by definition $P(c)$, and the way the building blocks are glued together is given by the maps between $P(c)$ and the images of other objects by $P$.
Now what is with representables? Well the representable $y(c)$ really is the presheaf freely generated by one building block of type $c$. The thing is a building block of type $c$ might not function all on its own, and you might need to throw in other things so that it can live, and that is what the representable presheaves are about. In my graph examples, there are two representable presheaves, which are $y(0)$ - the "single vertex graph" - and $y(1)$ - the "single arrow graph", composed with two vertices and an arrow between them.
Now I need to say a word about Yoneda lemma. The Yoneda lemma states that you can find the building blocks of a presheaf by looking at how it relates to the representable presheaves, instead of looking at the sets defined by the presheaves. More precisely, it says that $P(c)$ is the same as the morphisms of presheaves from $y(c)$ to $P$. In my graph examples, itsays that a vertex in the graph is the same as a morphism from the one vertex graph to my graph, and an arrow is the same as a morphism from the one arrow graph to my graph.
This is good, because it lets us assemble building blocks in the category of presheaves (as a colimit). More precisely, the set of all building blocks of $P$, which is given by $\bigsqcup_{c\in\textbf{C}} P(c)$, can be seen directly in the category of presheaves as the colimit $\coprod_\limits{c\in\textbf{C}\\ p\in P(c)} y(c)$. Then you need to take a coequalizer because $P$ is not just a bunch of building blocks thrown in together, but they are glued onto one another. The same kind of reasoning shows you that the coequalizer exactly explains to you how to glue these together. I let you carry it on explicitly if you want to.
Now to prove what you want to prove, you have to make all this story a bit more formal, but there is essentially one ingredient : the Yoneda lemma. Using this lemma, you should be able to prove directly that $P$ is the coequalizer that you gave (remember that colimits are computed pointwie in a presheaf category, and that you know very well how to compute them in sets) 
