Let us begin by describing the metric on $\widetilde{SL(2,\mathbb{R})}$.
Let on the tangent space $T\mathbb{H}^2$ of the hyperbolic plane there is a natural metric, called the Sasaki metric, built on the following way: for each $p=(x,v)$ we have a surface $S\subset T\mathbb{H}^2$ given by all possible parallel transports of $v$. We have a decomposition of $T_{p}\mathbb{H}^2$ as a sum $T_pS\oplus T_x\mathbb{H}^2\cong T_{p}\mathbb{H}^2$, on which we can put the metric $g_x\oplus g_x$, where $g_x$ is the hyperbolic metric on $T_x\mathbb{H}^2$. The unit tangent bundle $U\mathbb{H}^2\subset T\mathbb{H}^2$ inherits a riemannian metric, and so does its universal cover $\widetilde{U\mathbb{H}^2}\cong\widetilde{SL(2,\mathbb{R})}$.
With this metric $X=\widetilde{SL(2,\mathbb{R})}$ is the Thurton geometry also known as $\mathbb{H}^2\widetilde{\times}\mathbb{E}$. It is a line bundle over $\mathbb{H}^2$ whose connection has curvature $1$, and its isometry group is given by the following short exact sequence:$$1\to\operatorname{Isom}(\mathbb{R})\to\operatorname{Isom}(X)\to\operatorname{Isom}(\mathbb{H}^2)\to1$$ The group $G=\operatorname{Isom}(X)$ is a lie group of dimension four with two connected components corresponding to the (resp. non-)orientable isometries of $\mathbb{H}^2$.
Question: Why do we have this short exact sequence, i.e. why this is the right isometry group? While writing down what I understand I'll pose extra questions which will make this first broad question more specific.
A Lie group-theoretic approach to this question is the following: the space $X$ is a Lie group with a left-invariant metric, so it is itself a subgroup of is isometries. Since it is the universal cover of $SL(2,\mathbb{R})$ which is a $2$-fold cover of $PSL(2,\mathbb{R})$ we get a short exact sequence $$1\to \mathbb{Z}\to X\to PSL(2,\mathbb{R})\to 1$$ Now, it is well known that $PSL(2,\mathbb{R})\cong\operatorname{Isom}^+(\mathbb{H}^2)$. When we look at the action by isometries of $\mathbb{Z}\subset X$ on $X$ we see that it just translate each fiber by an integer, so it is clear that we can extend the action of $\mathbb{Z}$ to an action of $\mathbb{R}\cong\operatorname{Isom}(\mathbb{R})$ which now contains also non-integer translations. Hence get a short exact sequence $$1\to\operatorname{Isom}(\mathbb{R})\to\Gamma\to\operatorname{Isom}^+(\mathbb{H}^2)\to1$$ where $\Gamma$ is a subgroup of $\operatorname{Isom}(X)$. So we only need to answer the two following questions: how do we see that we also have isometries of $X$ that projects to non-orientable isometries of $\mathbb{H}^2$? How do we see that we do not have any other isometry?
A geometric approach could be the following. It is true that the bundle $U\mathbb{H}^2$, and hence $X$, is totally geodesic, furthermore the fibers of $U\mathbb{H}^2$ are circles, while the base space is contractible. Is this enough to deduce the following short exact sequence? $$1\to\operatorname{Isom}(S^1)\to\operatorname{Isom}(U\mathbb{H}^2)\to\operatorname{Isom}(\mathbb{H}^2)\to1$$