Confused about the Arrow Category If we use the definition of the Arrow category and the notation from here.
$$
\require{AMScd}
\begin{CD}
A @>h>> C \\
@VVfV @VVgV \\
B @>k>> D 
\end{CD}
$$
I think I can understand how the object $f$ from the $C^2$ category gets transformed using $h$ and $k$ into object $g$ if the starting category is just a directed graph (is quiver the correct name?). One just takes $h(A)$ as start and $k(B)$ as the end of the new arrow $g$.
But I'm confused when we start from the $\mathsf{Set}$ category. That is, if in our $C^2$ category the objects are functions. In particular, I don't see a way how construct the function $g\colon C \to D$ from only $f\colon A \to B,\ h\colon A \to C$ and $k\colon B \to D$? None of $f,g,h$ have $C$ as their domain, so there must be something else? Or have I misunderstood the types of $h,k$?

$C^{2}$ has objects that are arrows $f:a\to b$ of $C$. 
A morphism $\phi :f\to g$ is a pair $(h,k)$ such that $k\circ f=g\circ h$
$\tag 1 \begin{matrix}
\operatorname a & \xrightarrow{{f}} & \operatorname b \\
\left\downarrow h\vphantom{\int}\right. & & \left\downarrow k\vphantom{\int}\right.\\
\operatorname c&  \xrightarrow{g} & \operatorname d
\end{matrix}$
You can check that this gives a bonafide category. 
But the notation $C^{2}$ tells us why the category, in some sense, must be defined as it is: 
A naive way of looking at this would be to say that $C^{2}$ has objects that are $\textbf {functors}$  $F:2\to C$ where $2=\left \{ \bullet, \bullet \bullet \right \}$ is the category with two distinct objects and exactly one arrow $*:\bullet \to \bullet  \bullet $
$F$ will then be specified by its action on $\bullet $ and $\bullet \bullet $:
$F(\bullet )=a$ and $F(\bullet \bullet)=b$ 
and its action on $*$:
$F(*):F(\bullet )\to F(\bullet \bullet)$ which is just an arrow $a\overset{f}{\rightarrow} b$.
Now consider two objects (functors), $F$ and $G$, in $C^{2}$. 
An arrow $\tau :F\to G$ will be a natural transformation of the functors $F$ and $G$, and is specified by its components, $\tau _{\bullet}$ and $\tau _{\bullet \bullet}$ so in fact $\tau $ is a pair $(\tau _{\bullet}, \tau _{\bullet \bullet})$.
The corresponding naturality square is then
$\tag 2 \begin{matrix}
\operatorname F(\bullet ) & \xrightarrow{{F(*)}} & \operatorname F(\bullet \bullet )  \\
\left\downarrow \tau _{\bullet}\vphantom{\int}\right. & & \left\downarrow \tau _{\bullet \bullet}\vphantom{\int}\right.\\
\operatorname G(\bullet )&  \xrightarrow{G(*)} & \operatorname G(\bullet \bullet )
\end{matrix}$.
Now, to finish, set $(\tau _{\bullet}, \tau _{\bullet \bullet})=(h,k)$ and recall that $F(*)=f$ and $G(*)=g$ to see that $(1)$ and $(2)$ are the same.
We conclude that our original definition of the arrows in $C^{2}$ is the right one.
