How to get the Bertotti-Robinson metric into its conformally flat version? The Bertotti-Robinson metric is a very nice exact solution to the Einstein-Maxwell field equations in vacuum.  It is usually stated into these equivalent versions (where $r = \sqrt{x^2 + y^2}$ and $\bar{r} = \sqrt{\bar{x}^2 + \bar{y}^2 + \bar{z}^2}$ and $R$ is just a constant):
\begin{gather}
ds^2 = dt^2 - \frac{1}{(1 + r^2 / 4 R^2)^2} (dx^2 + dy^2) - \sin^2 {\!(t/R)} \, dz^2, \tag{1} \\[12pt]
ds^2 = dt^2 - R^2 (d\vartheta^2 + \sin^2 \vartheta \: d\varphi^2) - \sin^2 {\!(t/R)} \, dz^2, \tag{2} \\[12pt]
ds^2 = (1 + \tilde{z}^2 / R^2) \, d\tilde{t}^2 - R^2 (d\vartheta^2 + \sin^2 \vartheta \: d\varphi^2) - \frac{1}{(1 + \tilde{z}^2 / R^2)} \, d\tilde{z}^2, \tag{3} \\[12pt]
ds^2 = \frac{R^2}{\bar{r}^2} (d\eta^2 - d\bar{x}^2 - d\bar{y}^2 - d\bar{z}^2). \tag{4}
\end{gather}
While I know how to find the coordinates transformations that change (1)-(2)-(3) into each other, I'm unable to find the coordinates transformations that give version (4) (the conformally flat version).
So how to get metric (4) from (1), or (2), or (3) ?
 A: If one applies the standard coordinate transformation from 3D Cartesian coordinates $(\bar{x},\, \bar{y},\, \bar{z})$ in (4) to spherical coordinates $(r,\theta,\phi)$, then the metric (4) becomes:
$$
 ds^2= \frac{R^2}{\xi^2}(d\eta^2-d \xi^2)-R^2 (d\theta^2 + \sin^2 \theta d\phi^2),
$$
where we renamed the radial coordinate to $\xi$ to underscore that it no longer enters into the spherical part. This forms explicitly demonstrates the $AdS_2 \times S_2$ structure of the solution and has the same “angular” part as metrics (2) and (3).
So, what is left is the coordinate tranformation of the $AdS_2$ factor, which should not pose a problem, since the metric $\xi^{-2}(d\eta^2-d \xi^2)$ is simply a Wick-rotated standard metric of Poincaré half-plane model of the hyperbolic plane.
A: I found a solution to my question.  I consider A.V.S.'s answer (thanks dude!):
(4) can be written this way, using spherical coordinates, with $\bar{r} = \sqrt{\bar{x}^2 + \bar{y}^2 + \bar{z}^2}$:
\begin{align}
ds^2 &= \frac{R^2}{\bar{r}^2} \, d\eta^2 - \frac{R^2}{\bar{r}^2} \, \big( \, d\bar{r}^2 + \bar{r}^2 \, (d\vartheta^2 + \sin^2 \! \vartheta \: d\varphi^2) \big) \\[12pt]
&= \frac{R^2}{\bar{r}^2} \, (d\eta^2 - d\bar{r}^2) - R^2 \, (d\vartheta^2 + \sin^2 \! \vartheta \: d\varphi^2). \tag{5}
\end{align}
The angular part is identical as the one appearing in (2).  Now, I need to get this "radial" part:
\begin{equation}\tag{6}
ds_{\text{rad}}^2 = dt^2 - \sin^2 \! (t / R) \, dz^2.
\end{equation}
I'll substitute the following coordinates transformation into (5), where $F(z / R)$ is some function to find:
\begin{align}
\eta &= R \, F(z / R) \sinh{(\sigma / R)}, \\[12pt]
\bar{r} &= R \, F(z / R) \cosh{(\sigma / R)}.
\end{align}
Then the "radial" part of (5) becomes:
\begin{equation}\tag{7}
ds_{\text{rad}}^2 = \frac{1}{\cosh^2 {\! (\sigma / R)}} \, d\sigma^2 - \frac{F^{\prime \, 2}}{F^2 \cosh^2 {\! (\sigma / R)}} \, dz^2.
\end{equation}
To get rid of the pesky factor $F^{\prime \, 2} / F^2$, I'll use $F(z / R) = e^{z / R}$ :
\begin{equation}\tag{8}
ds_{\text{rad}}^2 = \frac{1}{\cosh^2 {\! (\sigma / R)}} \, d\sigma^2 - \frac{1}{\cosh^2 {\! (\sigma / R)}} \, dz^2.
\end{equation}
Now, I introduce a new time variable:
\begin{equation}\tag{9}
d\theta = \frac{1}{\cosh{\! (\sigma / R)}} \, d\sigma \qquad \Rightarrow \qquad \theta = 2 R \arctan{\big(\tanh{(\sigma / 2 R)} \big)}.
\end{equation}
Some algebra gives $\cosh{(\sigma / R)} = 1/\cos{(\theta / R)}$.  Then (8) becomes:
\begin{align}
ds_{\text{rad}}^2 &= d\theta^2 - \cos^2{\! (\theta / R)} \, dz^2 \\[12pt]
&\equiv d\theta^2 - \sin^2 {\! (\theta / R + \pi / 2)} \, dz^2. \tag{10}
\end{align}
Finally, I define $t = \theta + \pi R / 2$, so I recover (6).  So the complete transformation that take (4) and gives (2) is this:
\begin{align}
\eta &= R \, \frac{\cos{(t / R)}}{\sin{(t / R)}} \, e^{z / R}, \\[12pt]
\bar{r} &= R \, \frac{1}{\sin{(t / R)}} \, e^{z / R}.
\end{align}
This transformation covers only a part of spacetime since
\begin{align}
\frac{\eta}{\bar{r}} &= \cos{(t / R)},
& \eta^2 - \bar{r}^2 &= -\, R^2 \, e^{2 z / R} < 0.
\end{align}
