Well, the relevant line in Ahlfohrs just says "For two-sided arcs the same will be true with obvious modifications. This is the second edition (1966), chapter 6, section 1.3 "Use of the Reflection principle," pages 225-226, just after the Riemann Mapping Theorem.
So, take your domain (you need to type in something about $r e^{i \theta}$ with $0 \leq r < 1$). First, take the principal branch of the square root. That maps your domain to the semicircle $|z| < 1, \; \mbox{Re}\; z > 0.$ Then we have a domain bounded by exactly two free one-sided analytic arcs. So, Theorem 4, section 1.4, page 227, both the boundary line segment and the boundary semicircle are mapped 1-1 and analytically to the boundary circle of the unit disk.
However, this says to me that the original slit does not have a single-valued map to the unit circle, it is evidently double-valued, just as the square root. Maybe it's just me.
EDIT: I have it correct. See pages 18-19 in Boundary Behavior of Conformal Maps by Christian Pommerenke. What he does is switch the order, the conformal map starts in the unit disk, denoted $$f : \mathbb D \rightarrow G.$$ Note: Falcao of Atletico Madrid just scored on Real Madrid, drawn 1-1 at minute 56. OK, the boundary of $\mathbb D$ is the unit circle, denoted $\mathbb T.$
Continuity Theorem. The function $f$ has a continuous extension to $\mathbb D \cup \mathbb T$ if and only if $\partial G$ is locally
connected.
Caratheodory Theorem. The function $f$ has a continuous and injective extension to $\mathbb D \cup \mathbb T$ if and only if
$\partial G$ is a Jordan curve.
If $\partial G$ is locally connected but not a Jordan curve then
parts of $\partial G$ are run through several times. Some of the
possibilities are indicated in Fig. 2.1 where points with the same
letters correspond to each other. Note that e.g. the arcs $a_2 d_1$
and $a_1 d_2$ are both mapped onto the segment $ad.$
Here's an outline; I'll leave the details to you:
The map you have will send the unit disc to a half plane. To get from a half plane to all of $\mathbb{C}$ minus a ray, postcompose with $z\mapsto z^2$. Now, to get the missing ray where you want it, rotate and translate.
Lastly, look at the pre-image of $0$. You can precompose with an automorphism of the disk sending $0$ to that point. Then all that's left is to check that, when you compose all these maps, the derivative is a positive number.
Best Answer
If you think of the domain $D=\mathbb C \setminus {(-\infty,1]\cup[1,\infty)}$, this is not a doubly slit domain on the sphere, but it only has a single slit, passing through infinity (note that $\infty \notin D$). So you will have to move this slit to a slit from $0$ to $\infty$. This can be done with a Mobius transformation that preserves the real line, and maps $1$ to $0$ and $-1$ to $\infty$, say. It is easy to see that $(z-1)/(z+1)$ works.
Now you can unfold $\mathbb C \setminus [0,\infty)$ using $z^{1/2}$ (the branch that maps $-1$ to $i$). The image is the upper half plane, and it is easy now to map it to the unit disk with the Cayley transform $(z-i)/(z+i)$.