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 $h,k$?
 A: $C^{2}$ 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.
A: Your mistake is trying to think of morphisms as transformations; morphisms in this category are simply commutative squares; they don't have to "do" anything, aside from satisfy the algebraic properties of a category.
Also, the pair $(h,k)$ is not necessarily sufficient to uniquely identify a morphism in this category: you may need more. e.g. the additional context that we are to construe $(h,k)$ as a morphism from $f$ to $g$.
On this point, an analogy may help; many people take "function" to mean its graph: a set of ordered pairs that satisfies the "vertical line test". If you do this, then specifying a function is not enough to uniquely identify a morphism of $\mathbf{Set}$. If you construe functions as morphisms, you can compute the domains and products of said morphisms using just the functions, but you can't determine the codomains of the morphisms from just the function.
