$\DeclareMathOperator\Sl{Sl}\DeclareMathOperator\PSl{PSl}\DeclareMathOperator\Isom{Isom}$Let $\widetilde{\Sl_2}$ be the Thurson geometry that can either be described as the universal cover of $\PSl(2,\mathbb{R})$, or as the twisted line bundle over the hyperbolic plane $\mathbb{H}^2\mathbin{\tilde{\times}}\mathbb{E}$.

Its isometry group $\Isom(\widetilde{\Sl_2})$ is a $4$-dimensional Lie group with $2$ connected components, whose identity component $\Gamma$ fits into a short exact sequence $$0\to \Isom(\mathbb{E})\overset{i}{\to} \Gamma\overset{\pi}{\to}\Isom^+(\mathbb{H}^2)\to 0.$$ In fact this is even a central extension since $\Isom(\mathbb{E})\cong\mathbb{R}\triangleleft\Gamma$ is a central normal subgroup. Notice that this s.e.s. does not split, otherwise the group $\Gamma$ would be a direct product and we would have the non-twisted geometry $\mathbb{H}^2\times\mathbb{R}$. (See for example [Scott - The geometries of 3-manifolds] for more details on Thurston geometries.)

Hence, by the general theory of central extensions, there exist a map $\Phi:\Isom^+(\mathbb{H}^2)^2\to\mathbb{R}$ such that $\Gamma$ can be described in the following way: $\Gamma\cong(\mathbb{R}\times\Isom^+(\mathbb{H})^2,\circ_\Phi)$ where the composition law is the following: $$(a,f)\circ_\Phi(b,g)=(a+b+\Phi(f,g),fg).$$ To give a little more details we can say that the map $\Phi$ measures how much the s.e.s. fails to split: if $c:\Isom^+(\mathbb{H}^2)\to\Gamma$ is a section (of sets!) of the projection $\pi$ then for two general elements $f,g\in\Isom^+(\mathbb{H}^2)$ the elements $c(f)c(g)$ and $c(fg)$ differ by a unique element $i(a)$, i.e. $c(f)c(g)=i(a)c(fg)$. Then $\Phi$ is defined to be $\Phi(f,g)=a$. (See for example [Brown - Cohomology of groups] IV.3 for more details about extensions with abelian kernel.)

I am trying to answer the following question: what is the function $\Phi$ that gives rise to $\Gamma$?

Here is my attempt. First of all let us be aware of the nice series of equivalences of Riemannian manifolds $$\Isom^+(\mathbb{H}^2)\cong \PSl(2,\mathbb{R})\cong U\mathbb{H}^2$$ where $U\mathbb{H}^2\subset T\mathbb{H}^2$ is the unitary bundle on the hyperbolic plane that inherits is metric as a Riemaniann submanifold of the tangent bundle endowed with the Sasaki metric. Note that this metric comes with a couple of nice features:

1. An isometry $f$ of $\mathbb{H}^2$ acts as an isometry of $T\mathbb{H}^2$ via the differential $df$, hence also on $U\mathbb{H}^2$ when restricted on it.
2. The fibers $S^1$ over each point are totally geodesic.

It follows in particular that $df$ sends fibers to fibers and acts on each of them as a rotation of an angle $\theta_x$ (is this angle constant w.r.t $x$? I think so, but wouldn't know how to prove it). Hence $df$ acts also on the universal cover $\widetilde{\Sl_2}$ sending each fiber to the corresponding one, and translating it by a length $\theta_x$. In this way we have a section $\Isom^+(\mathbb{H}^2)\to\Gamma$ defined by $f\to df$. Now let us take to isometries $f,g\in\Isom^+(\mathbb{H}^2)$ and look at the elements $df,dg,d(fg)^{-1}\in\Gamma$. Of course the projection of their composition acts as the identity on $\mathbb{H}$, so their composition acts as a translation on each fiber. If we choose a point $x\in\mathbb{H}^2$ their composition will act as a rotation of $S^1\cong U_x\mathbb{H}^2$ of angle $\theta_x(f,g)$, and it should be pretty straightforward to check that $\Phi(f,g)=\theta_x(f,g)$ is the function I'm looking for.

Here are some open points: first of all I strongly believe that that angle $\theta_x$ should not depend on $x$. This might simplify the calculation a little bit. Nevertheless the calculations still looks to me fairly annoying, as they involve a lot of nasty differentials in coordinates or parallel transports. So I wonder is there any other more straightforward way to find this (or another) explicit description? Maybe it could involve some Lie-algebra work that I am not aware of.