= Solution
Let $\xi$ denote <geodesic> distance from the charge; $R$ is the fixed curvature radius, not the radial variable. In round three-dimensional <hyperbolic space>, a sphere has area
$$
A(\xi)=4\pi R^2\sinh^2(\xi/R).
$$
Using the convention that <electric flux> is $Q/\epsilon_0$, <Gauss's law> gives the radial orthonormal <electric field>
$$
\boxed{\mathbf E(\xi)=\frac{Q}{4\pi\epsilon_0R^2\sinh^2(\xi/R)}\,\widehat{\boldsymbol\xi}.}
$$
Its magnitude is $|Q|/A\epsilon_0$, and a negative charge reverses the direction. In rationalized natural units set $\epsilon_0=1$; Gaussian conventions instead use flux $4\pi Q$.
For $\xi\ll R$ this approaches $Q/(4\pi\epsilon_0\xi^2)$. For $\xi\gg R$, $E\sim Qe^{-2\xi/R}/(\pi\epsilon_0R^2)$. This <hyperbolic-space Coulomb field> falls rapidly because the sphere area grows exponentially; conserved <electric flux> is unchanged. A potential vanishing at infinity is $V=Q[\coth(\xi/R)-1]/(4\pi\epsilon_0R)$.
The original PDF actually prints $\sin\theta$ rather than $\sin^2\theta$ in the angular <metric tensor>. The round angular factor used above is required by its description as <hyperbolic space>. If the printed angular <metric tensor> is taken literally, the sphere-area coefficient is instead
$$
\Omega_*=2\pi\int_0^\pi\sqrt{\sin\theta}\,d\theta
=2\pi\sqrt\pi\,\frac{\Gamma(3/4)}{\Gamma(5/4)},
$$
and a radial Gauss-law solution has $E=Q/[\epsilon_0\Omega_*R^2\sinh^2(\xi/R)]$. That angular geometry is not a smooth round sphere, so the literal <metric tensor> does not describe the asserted homogeneous <hyperbolic space>. Both readings retain the same radial area-growth factor; the geometric repair is explicit.
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