Ergodic Theory with a view towards Number Theory, Chapter 9, Page 281.
Lemma 9.2. The action of $\mathrm{PSL}_2(\mathbb{R})$ on $\mathrm{T}^1\mathbb{H}$ is simply transitive.
Notice that this allows us to describe the unit tangent bundle to $\mathbb{H}$ as $$\mathrm{T}^1\mathbb{H}\cong\mathrm{PSL}_2(\mathbb{R}).\tag{9.4}$$
In order to do this, we have to choose an arbitrary reference vector $(z_0,v_0)$ in $\mathrm{T}^1\mathbb{H}$ which corresponds to $I_2\in\mathrm{PSL}_2(\mathbb{R})$; the identification is then given by $g\mapsto\mathrm{D}g(z_0,v_0)$. We will make the convenient choice $z_0=\mathrm{i}$ and $v_0=\mathrm{i}$. That is, the reference vector is the upward unit vector based at the imaginary unit $\mathrm{i}\in\mathbb{H}$. Under the resulting identification the action of $\mathrm{PSL}_2(\mathbb{R})$ on $\mathbb{H}$ is conjugated to the action of $\mathrm{PSL}_2(\mathbb{R})$ by left multiplication on $\mathrm{PSL}_2(\mathbb{R})$.
How do we have that $$\mathrm{T^1}\mathbb{H} \cong\mathrm{PSL}_2(\mathbb{R})$$ And the last three lines in the attached image above (Under the resulting......$\mathrm{PSL}_2(\mathbb{R})$).
With the Iwasawa decomposition $SL_2\mathbb{R}=KAN$, where $K=SO(2)$, $A$ consists of diagonal elements, and $N$ consists of unitriangular elements, we can show $SL_2\mathbb{R}$ acts transitively on $\mathbb{H}$. Indeed, the subgroup $AN$ itself (upper triangular elements) acts transitively: you can use $a\in A$ to make the imaginary part of a point whatever you need it to be, then use $n\in N$ to make the real part whatever you want it to be.
Indeed, given any $(z,v)$ (with $v$ the unit tangent vector), you can use $g\in G$ to turn this into $(i,w)$ for some tangent vector $w$. Then, you can pick a $k\in K=SO(2)=\mathrm{Stab}(i)$ to apply to get $(i,i)$. Thus you can get $(i,i)$ from any $(z,v)$ (and thus also conversely).
The orbit-stabilizer theorem says if $G\curvearrowright\Omega$ is transitive then $\Omega\cong G/\mathrm{Stab}(\omega)$. In the case of a sharply transitive action (i.e. a regular action), this means $\Omega\cong G$ are equivalent $G$-sets. Picking the correspondence between $G$ and $\Omega$ amounts to picking a $\omega\in\Omega$ to correspond to $e\in G$. (In the case of $G$ a vector space and $\Omega$ an affine space, this means picking an "origin.") Then $g\mapsto g\omega$ is the equivalence $G\cong\Omega$. We have $G$ acting on itself by left-multiplication, of course.
Your source is choosing $\omega=(z_0,v_0)$ as a "reference vector" - specifically $(i,i)$ is nice.