We have a sequence of maps
$$f_i:\ D_{i-1}\to D_i,\quad z_{i-1}\to z_i\qquad (1\leq i\leq 5)\ .$$
Here $z_i$ does not denote a certain point in the $z$-plane, but the coordinate variable in the $i$th auxiliary complex plane. $Z_0:=\sqrt{2}-1$ is the $z_0$-coordinate of a certain point $Z$ we are interested in.
$D_0$ is the unit disk in the $z_0$-plane minus the points $z_0\leq0$. The map
$$f_1:\ z_0\mapsto z_1:={\rm pv}\sqrt{z_0}$$
maps $D_0$ onto the right half $D_1$ of the unit disk in the $z_1$-plane. Thereby the point $Z_0$ is mapped onto a point $Z_1\in\ ]0,1[\ $.
The Moebius map
$$f_2:\ z_1\mapsto z_2:=-{z_1-i\over z_1+i}$$
maps $i$ to $0$ and $-i$ to $\infty$. Furthermore $f_2(0)=1$, $\ f_2(1)=i$. From general properties of Moebius maps it then follows that $D_2:=f_2(D_1)$ is the first quadrant, and that $f_2$ maps the real axis onto the unit circle. Therefore $Z_2=f_2(Z_1)$ is a point between $1$ and $i$ on the unit circle.
The map
$$f_3:\ z_2\mapsto z_3:=z_2^2$$
maps the first quadrant $D_2$ onto the upper half-plane $D_3$, whereby $f_3(1)=1$, $\ f_3(i)=-1$, and the quarter unit circle in $D_2$ is mapped onto the upper half of the unit circle in $D_3$. Therefore the point $Z_3:=f_3(Z_2)$ is lying on this upper half of the unit circle, too.
The Moebius map
$$f_4:\ z_3\mapsto z_4:=i{z_3-i\over z_3+i}$$
maps the upper half plane $D_3$ onto the unit circle $D_4$. Thereby $f_4(-1)=-1$, $\ f_4(1)=1$, and the unit circle of the $z_3$-plane is mapped onto the real axis of the $z_4$-plane. It follows that $Z_4=f_4(Z_3)$ is a real number between $-1$ and $1$.
Doing the calculations $Z_4$ should simplify to an expression defining a real number $\alpha\in\ ]{-1},1[\ $. Letting
$$f_5:\ z_4\mapsto{z_4-\alpha\over 1-\alpha z_4}$$
you finally arrive at the required map
$$f:=f_5\circ f_4\circ f_3\circ f_2\circ f_1\ .$$
The first three steps from solution in a comment are fine:
Use the map $z\mapsto z^2$ to send $\{|z|<1 : \operatorname{Re} z>0\}\setminus [0,1/2]$ to $\{|z|<1\}\setminus [ā1,1/4]$.
Then use the map $z\mapsto \dfrac{zā1/4}{1āz/4}$ to send it to $\{|z|<1\}\setminus[ā1,0]$.
Use $z\mapsto \sqrt{z}$ to send it to the right half disk.
I think $\dfrac{z-i}{z+1}$ has a typo: should be $z+i$ in the denominator. Easily fixed. Anyway, my preference is to use the Joukowski map $z\mapsto z+z^{-1}$ for half-disks. It sends the upper half of the unit disk to the lower halfplane, and the lower half-disk to the upper halfplane. So, the remaining steps can be replaced with $z\mapsto -iz$ followed by $z\mapsto z+z^{-1}$.
since the map $z\mapsto z^2$ is one-to-one in the right half disk and all the other subsequent maps are also injective in their domain of use, hence the resulting composition map is also injective and therefore conformal.
This is correct.
What is usually the best strategy to construct maps from/to domains with slits?
Exactly what you used: a combination of square and square root maps that "push" the slit back into the boundary from which it sticks out.
Best Answer
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 $\mathbb T.$