So here's my attempt at why the homset functor \\(\hom : \mathcal{C}^\mathrm{op} \times \mathcal{C} \to \mathbf{Set}\\) has to be contravariant in the first argument. Suppose we attempt to define the homset functor as \\(\hom : \mathcal{C} \times \mathcal{C} \to \mathbf{Set}\\) such that any pair of objects \\((c,c')\\) is mapped to the set of morphisms \\(\mathcal{C}(c,c')\\) between these objects. We have to define what happens the morphisms \\( (f,f') : (c,c') \to (d,d') \\), and applying the homset functor to the morphism pair should give us a map sending morphisms in \\(\mathcal{C}(c,c')\\) to morphisms in \\(\mathcal{C}(d,d')\\).

Let \\(h \\) be any element of \\(\mathcal{C}(c,c')\\) and \\(h'\\) an element of \\(\mathcal{C}(d,d')\\) such that \\(\hom (f,f') (h)=h'\\). I drew the following diagram (I actually drew a bunch of arrows for each homset, but that's hard to do in LaTeX):

\[
\begin{matrix}
& & f' & & \\\\
& c' & \rightarrow & d' &\\\\
h & \uparrow & & \uparrow & h'\\\\
& c & \rightarrow & d &\\\\
& & f & & \\\\
\end{matrix}
\]

From this diagram, we get two compositions \\(f'h : c \to d'\\) and \\(h'f : c \to d' \\), so the morphism \\((f,f')\\) ends up sending the homset \\(\mathcal{C}(c,c')\\) to \\(\mathcal{C}(c,d')\\); not quite what we want. But it's easy to see that if the arrow \\(f\\) was turned around, we could precompose \\(h\\) with it.

In \\(\mathcal{C}^\mathrm{op}\\), \\(f : d \to c\\) as desired, so we get the following commutative diagram:

\[
\begin{matrix}
& & f' & & \\\\
& c' & \rightarrow & d' &\\\\
h & \uparrow & & \uparrow & h'\\\\
& c & \leftarrow & d &\\\\
& & f & & \\\\
\end{matrix}
\]
where \\(h' = f'hf\\), which looks a bit like conjugation in a group. So the functor \\(\hom\\) sends morphisms in \\(\mathcal{C}^\mathrm{op} \times \mathcal{C} \\) to functions between homsets defined by the previous formula.