Different setup for Pedersen Commitment

I have read many question on this website and understood the Pedersen commitment until I came across with this page.

This page, it computes $\mathcal h= g^s \bmod p$ where $s$ is secret, instead of using $h$ and $g$ as a generator of a group $G$ like in first page. Is there a specific reason for this?

In a prime order group, every element except for the identity is a generator. As such, $g^s \bmod p$ will be a generator except with probability $1/q$. This is just one way of ensuring that $h$ is a generator (that is true for generic groups). Note that given $s$, it is possible to decommit to any value. As such, $s$ behaves as a type of trapdoor. This means that the receiver (or some external party). has to choose $h$ and not the committer. In some protocols, this is used by having the receiver choose $h$ and prove that it knows the appropriate $s$ with a zero-knowledge proof of knowledge. This enables the simulator to extract $s$ and use it during the simulation. Of course, in a real execution, it remains secret since the proof is zero knowledge.