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
$$f:=f_5\circ f_4\circ f_3\circ f_2\circ f_1\ .$$
Best Answer
There is a derivation of the bijective Möbius transformations $\mathbb{D} \to \mathbb{D}$, the non bijective ones are easily found from this
Without loss of generality, you can look at $$g(z) = C\frac{z-A}{z-B}, \qquad A \in (0,1),\ |B| > 1,\ C > 0$$
$f(z) = \frac{az+b}{cz+d} = \frac{a}{c}\frac{z+b/a}{z+d/c}$ and $g(z) = e^{-i\text{arg} C} f(-ze^{i\text{arg}(b/a)})$ has the form $C\frac{z-A}{z-B}, A > 0, C > 0$. Since $g(z)$ is bijective $\mathbb{D} \to \mathbb{D}$, the point where $g(\rho) = 0$ must be inside $\mathbb{D}$, so $A \in \mathbb{D} \implies A \in (0,1)$. Finally $|B| > 1$ since $g(z)$ is holomorphic on $\mathbb{D}$ and $B \ne A$.
By the maximum modulus principle (not hard to prove) you have that $g$ maps $\partial \mathbb{D} = \{|z|=1\}$ to itself, so that $$1 = |g(\pm i)| = |C| \frac{|\pm i-A|}{|i-B|} = |C| \frac{\sqrt{1+A^2}}{|\pm i-B|} $$ i.e. $|B-i| = |B+i| \implies B \in \mathbb{R}$,
and $1 = |g(1)| = C \frac{1-A}{|B-1|}= |g(-1)| = C \frac{1+A}{|B+1|}$ means $|B-1| < |B+1|$ so that $B > 0 \implies B > 1$.
Finally since $A \in (0,1),B > 1, C > 0$, you have $g(1) < 0, g(-1) > 0$ and $|g(1)| = |g(-1)| =1 $ means $$g(1) =C \frac{1-A}{1-B} =-1,\qquad g(-1) = C\frac{-1-A}{-1-B} = 1$$ i.e. $\frac{1-A}{B-1} = \frac{1+A}{B+1} \implies (1-A)(B+1) = (1+A)(B-1) \implies AB = 1$ and $f(-1) = 1 \implies C = \frac{1+1/A}{1+A} = 1/A$.
Going back to $f(z)$, you get the desired general form $$f(z) = e^{i \theta}\frac{z-a}{1-\overline{a}z}, \qquad |a| < 1$$