Why the mapping $1/\bar{z}$ look like the field lines of an electric dipole?

To get circles algebraically from a dipole, we need to use an inverse law, as holds in the plane, rather than an inverse-square law. Things are pleasantly symmetric if we put the charges at the complex numbers $\pm1$. The potential is $$ V(h, k) = \log\frac{(h + 1)^{2} + k^{2}}{(h - 1)^{2} + k^{2}}, $$ whose lines of force (flow lines of the gradient) are circles through $(\pm1, 0)$ and whose equipotentials are the family of orthogonal circles.

As for the Riemann sphere pictures, here are lines through the origin and the orthogonal family of concentric circles, mapped to the sphere by stereographic projection, and then mapped back to the plane as the sphere rotates about $\pm i$:

When either $\pm1$ is carried to the north pole (the point at infinity) by rotation, the field lines become radial rays through the origin.

Here are the curves seen from directly above (with the real axis vertical), without the sphere:

I feel somewhat as though I am just repeating what you yourself said in your post, so maybe this will not be helpful, but here goes anyway. One of the important properties of a Mobius transformation is that it maps lines and circles to lines and circles. In fact it is helpful to think of lines as circles which pass through the point infinity. This can be understood a bit more cleanly by working with on the Riemann sphere, but the intuition that a line is a "circle of infinite radius" should be perfectly serviceable for the moment.

The map $z \mapsto \frac{1}{\overline z}$ is an involution, that is applying it twice gives the identity map. Note that it fixes $i$ and it switches $0$ and $\infty$. In red you have drawn the set of all lines containing the point $i$. Or, from another perspective, in red you have draw the set of all "circles" containing the points $i$ and $\infty$. After you apply the transformation, you therefore get the set of all circles containing the points $i$ and $0$.

So one explanation of the picture is simply that if you draw the collection of all circles which contain any pair of distinct points $p,q \in \mathbb{C}$, then you get a picture resembling a magnetic dipole.

Real electric/magnetic dipoles do not have perfectly circular field lines like here; their shape depends on the field strength. Nevertheless, since the input is a configuration of lines that is invariant under conjugation, it suffices to consider only $\frac1z$ and not $\frac1{\overline z}$.

Consider an arbitrary line passing through $(1,0)$; its polar equation is $r=\frac{\cos a}{\cos(\theta-a)}$ where $a$ is the line's normal. The $\frac1z$ map then sends $$\left(\frac{\cos a}{\cos(\theta-a)},\theta\right)\to\left(\frac{\cos(\theta-a)}{\cos a},-\theta\right)$$ and so the equation of the transformed line is $r=\frac{\cos(\theta+a)}{\cos a}$, a circle containing the points $0$ and $1$. Varying $a$ runs through all possible circles containing $0$ and $1$, generating the observed dipole-like pattern.