Why are the morphisms in comma categories not defined the other way around? A comma category of functors $S:\mathcal A\to \mathcal C$ and $T:\mathcal B\to \mathcal C$ is a category where objects are pairs $(A,B,h)$ with $h:S(A)\to T(B)$, and morphisms are pairs $(f,g):(A,B,h)\to(A’,B’,h’)$, where the following diagram commutes:

Why isn’t this defined so that $T(g)$ is a morphism from $T(B’)\to T(B)$ instead? This intuitively seems more reasonable to me, since it would mean that we can interpret morphisms in the comma category as “turning $h$ into $h’$”. I am just now learning about comma categories so I don’t really have an intuition for them or for why they are used.
 A: Lots of constructions in category theory involve a commuting square that 'flows in one direction'. So in your picture, everything flows to the bottom right corner. Somehow this is the 'natural' thing to consider.
One particular example that may interest you is that of natural transformations (nLab, Wikipedia). Given functors $F, G: \mathcal{C} \to \mathcal{D}$, a natural transformation $\eta: F \Rightarrow G$ is essentially a morphism of functors. It is defined as follows: $\eta$ is a collection of arrows in $\mathcal{D}$ of the form $\eta_C: F(C) \to G(C)$ for each object $C$ in $\mathcal{C}$. These arrows should satisfy some naturality condition, namely that for every arrow $f: C \to C'$ in $\mathcal{C}$ we have that
$$
\require{AMScd}
\begin{CD}
F(C) @> F(f) >> F(C')\\
@V \eta_C VV @VV \eta_{C'} V \\
G(C) @>> G(f) > G(C')
\end{CD}
$$
commutes. So this is saying that $\eta$ 'is compatible with arrows in $\mathcal{C}$'.
You can probably already see the similarity with the diagram you provided. We can actually make that precise, we can reformulate the definition of a natural transformation in terms of a comma category. A natural transformation $\eta: F \Rightarrow G$ then corresponds to a functor $T: \mathcal{C} \to (F \downarrow G)$ such that $T(C) = (C, C, \eta_C)$ and $T(f) = (f, f)$. Here $(F \downarrow G)$  denotes the comma category. It might be nice to write this out for yourself to see why this is true.
