I will give two easier-seeming facts and proofs, and then show that the "colimit of representables" idea is obtained by the essentially same arguments in a more-general setting.
If $R$ is a ring, then any $R$-module $M$ is a quotient of a free module.
Proof: Choose generators for $M$ by some intricate procedure, or by taking every element of $M$ to be a generator. For each generator $m$, build a map $R^1 \to M$ specified by sending $1 \mapsto m$. This assignment extends to at most one map of modules since $1$ generates $R^1$ as a module, and moreover, it does extend, since the formula $r \mapsto mr$ constructs a valid map of right modules.
By summing these maps over the generating set, we obtain a map from a large free module to $M$. Every generator is in the image of this map since it is hit by its corresponding basis vector. Since the image is a submodule of $M$ that contains a generating set, the map is a surjection, and so $M$ is a quotient of a free module.
Any $R$-module $M$ has a presentation by generators and relations.
Proof: Witness $M$ as a quotient of a free module, which we will call the module of generators. The kernel of the quotient map is again an $R$-module (actually, it is a submodule of the module of generators), and so is itself a quotient of a free module (the module of relations). There is a natural map from the module of relations to the module of generators given by the quotient map followed by the inclusion, and the cokernel of this map is $M$. This shows that $M$ has a presentation by generators and relations.
Ok, and here are the corresponding facts and proofs for categories. Set $\mathcal{D} = \mathcal{C}^{op}$. By an "element" of $F \colon \mathcal{D} \to \mathrm{Set}$ we mean a pair $(d \in \mathcal{D}, x \in Fd)$. Essentially, an "element of F" is an element of the disjoint union of the various $Fd$. If $f \colon d \to d'$ is a morphism of $\mathcal{D}$, then we write $(d,x) \cdot f = (d', (Ff)(x))$. This gives a right action of $\mathcal{D}$ on the elements of $F$.
If $\mathcal{D} = C^{op}$ is a category, and $F : \mathcal{D} \to \mathrm{Set}$ is a functor to sets, then $F$ is a quotient of a disjoint union of representable functors.
Proof: Choose enough elements of $F$ so that every other element may be obtained from these by use of the action of $\mathcal{D}$. For example, one may choose every element of $F$ at this stage. Call this collection of elements the "generators". Then, for each generator $(d, x)$, build a map
$$ \mathcal{D}(d,-) \to F $$
specified by $(d, 1_d) \mapsto (d, x)$ and therefore, by naturality we would have at a general element $(d', f \colon d \to d') \mapsto (d,x) \cdot f$. As this formula does give a natural transformation, the desired map exists.
By taking the coproduct of these maps over the generating set, we obtain a map from a large disjoint union of representables to $F$. Every generator $(d,x)$ is in the image of this map since it is hit by the element $(d, 1_d)$ in the corresponding summand of the disjoint union. Since the image is a subfunctor that contains a generating set, it is a surjection, and so $F$ is a quotient of a disjoint union of representables.
Any functor is a the coequalizer of a map from one disjoint union of representables to another.
Proof: By the lemma, write $F$ as a quotient of a disjoint union of representables $X$, so that $X$ the $\mathcal{D}$-set of generators. The set of pairs of elements of $X$ that map to the same element of $F$ gives a subfunctor of $X \times X$. Write this subfunctor as a quotient of a disjoint union of representables $Y$, the relations. We obtain two natural transformations $Y \to X$ by the quotient map followed by each projection. The coequalizer of these two maps identifies all pairs of elements of $X$ that map to the same thing in $F$, so the coequalizer is isomorphic to $F$.
Remarks
For a good example to work by hand, I recommend setting $R=\mathbb{Z}[x,y]$ to be a graded ring with $|x|=|y|=1$, and taking $M$ to be a monomial ideal. The objects of $\mathcal{D}$ are the grades of $R$, and there can be some nice combinatorics in the construction of the relations.
In commutative algebra, it is often useful to extend a presentation to a free resolution. The analog would be finding a simplicial resolution by coproducts of representables.