Conceptual understanding of the global triviality of $SL(2,\Bbb{R})$ with typical fiber $S^1$ In Saunders' book on jet bundles, the following example is given:

Let $H=\{z\in\Bbb{C}\,:\,\text{im}(z)>0\}$ be the upper half plane regarded as a 2-dimensional real manifold, and consider the mapping $\pi:SL(2,\Bbb{R})\to H$ defined as
\begin{align}
\pi
\begin{pmatrix}
a&b\\
c&d
\end{pmatrix}&=
\frac{ai+b}{ci+d}
\end{align}
Then, $\pi$ is a surjective submersion, hence $\left(SL(2,\Bbb{R}),\pi,H\right)$ is a fibered manifold. Also, if we define $\rho:SL(2,\Bbb{R})\to S^1$ as
\begin{align}
\rho
\begin{pmatrix}
a&b\\
c&d
\end{pmatrix}&=
\left(\frac{a}{\sqrt{a^2+c^2}},\frac{c}{\sqrt{a^2+c^2}}\right),
\end{align}
then the mapping $SL(2,\Bbb{R})\to H\times S^1$, defined as $A\mapsto (\pi(A),\rho(A))$ is a global trivialization.

I managed to verify the claims made here (that $\pi$ is a surjective submersion and $(\pi(\cdot),\rho(\cdot))$ is indeed a diffeomorphism), but unfortunately, my proof was a direct but very tedious computation. I was hoping if someone could explain to me conceptually/geometrically why the mappings are chosen as such. For instance, I know can vaguely recognize (but can't fully make the connection) that the definition of $\pi$ has something to do with Mobius transformations in the complex plane, but I don't understand what is so special about $SL(2,\Bbb{R})$ here.
So, my questions are what is the significance of these maps, and why one ought to expect $\pi$ to be surjective submersion and $(\pi,\rho)$ to be a diffeomorphism, and hopefully someone can elaborate on the  "deeper meaning" of this example.
 A: A group $G$ is called a knit product of subgroups $H$ and $K$ if $G=HK$ and $H\cap K=1$. That is, if every group element $g\in G$ is uniquely expressible as $g=hk$ with $h\in H,k\in K$. When these are all Lie groups, that means $G\approx H\times K$ are diffeomorphic. Indeed, $(h,k)\mapsto hk$ is a diffeomorphism $H\times K\to G$. (Or if they're topological groups, this is a homeomorphism.)
Suppose we have a group action $G\curvearrowright\Omega$. The orbit-stabilizer theorem says every orbit $\mathrm{Orb}(\omega)$ is equivalent, as a $G$-set, to the coset space $G/\mathrm{Stab}(\omega)$. The way to think about this is that $G$ "covers" the orbit $\mathrm{Orb}(\omega)$, with the fibers (or "sheets" which cover each individual element) the cosets of $\mathrm{Stab}(\omega)$. The actual map we're thinking of as a covering is simply the evaluation map $G\to\mathrm{Orb}(\omega)$ given by $g\mapsto g\omega$, which restricts to a constant map $g\mathrm{Stab}(\omega)\mapsto g\omega$ on each fiber (coset). This is map intertwines with the $G$-action, and induces an equivariant bijection between $G/\mathrm{Stab}(\omega)$ and $\mathrm{Orb}(\omega)$.
If $G$ is a Lie group (maybe topological group), that means there is a fiber bundle
$$ \mathrm{Stab}(\omega)\to G\to \mathrm{Orb}(\omega). $$
In other words, in the general case $G$ is like a "twisted" version of $\mathrm{Orb}(\omega)\times\mathrm{Stab}(\omega)$.
Exercise. Prove if $H$ is a transitive subgroup and $K$ a stabilizer then $G=HK$ is knit.

The group $\mathrm{GL}_2\mathbb{C}$ acts on $\mathbb{C}^2$ by multiplication. That means $\mathrm{PGL}_2\mathbb{C}=\mathrm{GL}_2\mathbb{C}/Z(\mathrm{GL}_2\mathbb{C})$ (where the center $Z$ consists of the scalar matrices $\mathbb{C}^\times I_2$) acts on the complex pojective line $\mathbb{CP}^1$, which is the set (really, the topological space) of complex 1D subspaces of $\mathbb{C}^2$. Every 1D subspace is of the form $\mathbb{C}[\begin{smallmatrix}x\\y\end{smallmatrix}]$, and there is a bijection $\mathbb{CP}^1\to\mathbb{C}\sqcup\{\infty\}$ given by $\mathbb{C}[\begin{smallmatrix}x\\y\end{smallmatrix}]\mapsto\frac{x}{y}$. We can do "transport of structure," where we transport the action of $\mathrm{PGL}_2\mathbb{C}$ from $\mathbb{CP}^1$ to $\mathbb{C}\sqcup\{\infty\}$ - this turns out to be Mobius transformations!
The subgroup $\mathrm{PGL}_2\mathbb{R}$ preserves the subset $\mathbb{RP}^1\subset\mathbb{CP}^1$, or equivalently $\mathbb{R}\sqcup\{\infty\}\subset\mathbb{C}\sqcup\{\infty\}$. We may instead consider this an action of $\mathrm{SL}_2\mathbb{R}$. This has three orbits: the extended real number line and the upper and lower halves of $\mathbb{C}$ it separates. When $\mathrm{SL}_2\mathbb{R}$ acts on the upper half plane $H$, we may calculate the stabilizer $\mathrm{Stab}(i)=\mathrm{SO}(2)$. Also the Mobius transformations $\frac{az+b}{cz+b}$ have a nice subgroup of affine transformations $az+b$, which can further be decomposed as a knit product of homotheties $z\mapsto az$ and translations $z\mapsto b$. Thus, $G=\mathrm{SL}_2\mathbb{R}$ is a three-fold knit product of
$$ K=\left\{\begin{bmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \phantom{-}\cos\theta\end{bmatrix}\right\}, \quad A=\left\{\begin{bmatrix} \sqrt{a} & 0 \\ 0 & 1/\sqrt{a}\end{bmatrix}\right\}, \quad N=\left\{\begin{bmatrix}1 & b \\ 0 & 1\end{bmatrix}\right\} $$
The subgroup $B=AN$ (which is in fact a semidirect product) acts transitively on $H$. To construct an affine transformation from one point to another, first pick the unique homothety turns the one imaginary part into the other, then pick the unique translation that finishes the job. Thus, $G$ is a knit product $KB$.
In fact, we can find a $K$-equivariant projection $G\to K$. First, observe how $K$ acts on  $G$: to multiply out matrices $XY$, we essentially apply $X$ to the columns of $Y$. What kind of column vectors are possible for, say, the first column? Any nonzero vector, which $K$ acts on by rotations. If we normalize such vectors we get elements of the unit circle which is equivalent to $K$. This yields the desire projection $G\to S^1$.
Your projection $G\to H$ is essentially the "evaluation map" discussed in the context of the orbit-stabilizer theorem, the "covering." We expect its fibers to be cosets of $K$. Indeed, we can make this projection $B$-equivariant if we want, and $B$ acts regularly on $H$ so $B\approx H$ are equivalent as $B$-sets.
