Conformal mapping from the unit circle to the connected region

Such a bilinear (or Möbius) transformation will always map circles (such as $\partial D$) to circles (or lines), but the boundary $\partial \Omega$ of $\Omega$ is lens-shaped, with two $60^\circ$ vertices at $-1$ and $1$. So you do need something more.

I would start at $\Omega$, use a bilinear map to transport one of the vertices (e.g., $-1$) to infinity ($z\mapsto \frac1{z+1}$); this turns the two circular arcs of $\partial \Omega$ into lines intersecting at still the same angle of $60^\circ$ at the image point of the other vertex $1$, i.e., at $\frac12$. Now transport this to the origin ($z\mapsto z-\frac12$) and raise to a suitable power to turn the 60° into "smooth" 180° ($z\mapsto {z^3}$). In total, this turns $\Omega$ into a half plane, from where you probably know how to arrive at the unit disk.

In total, thus gives you $\Omega \to D$. To obtain $D\to \Omega$, try to invert the function.