The composition of conformal maps is conformal, so to obtain a conformal map between two domains, we can - if it seems more simple - map the first domain conformally to a simpler intermediate domain, and then map that intermediate domain conformally to the target (perhaps with more intermediate steps).
One needs to know some standard conformal maps of course. A well-known family of conformal maps are the Möbius transformations. These allow us to map any (open) haf-plane conformally to any (open) disk, mapping any prescribed point in the interior of the half-plane to the centre of the disk.
Knowing that, we need a conformal map from the quadrant to a half-plane. The boundary of the quadrant has a vertex where the two straight half-lines making up the boundary meet at a right angle. The boundary of a half-plane has no vertex, in the plane, it is a straight line, in the sphere, a circle. So we need something that straightens the right angle of the boundary of the quadrant.
Now one should remember that the power maps $z \mapsto z^k$ multiply angles at $0$ by $k$ - writing $z = \rho e^{i\varphi}$, we have $z^k = \rho^k e^{ik\varphi}$ - so to straighten the right angle at $0$ of the boundary of the quadrant, we need $k = 2$ (generally, to straighten an angle $\alpha$, we need the power $\pi/\alpha$ [which need not be an integer]). So the first part of our map is
$$s \colon Q \to \mathbb{H};\quad z \mapsto z^2$$
that maps the first quadrant $Q = \{ x+iy \in \mathbb{C} : x > 0, y > 0\}$ conformally to the upper half-plane $\mathbb{H} = \{ z \in \mathbb{C} : \operatorname{Im} z > 0\}$.
Then we need a conformal map $T \colon \mathbb{H} \to \mathbb{D}$ from the upper half-plane to the unit disk, that maps $s(1+i) = (1+i)^2 = 2i$ to $0$ and $s(0) = 0$ to $i$.
A Möbius transformation mapping $2i$ to $0$ and the real line (the boundary of the upper half-plane) to the unit circle is
$$T_0 \colon z \mapsto \frac{z-2i}{z+2i}.$$
That does not yet quite do what we want, since $T_0(0) = \frac{-2i}{2i} = -1$, so we compose it with a rotation that takes $-1$ to $i$, and that is multiplication by $-i$, so we get
$$T\colon z \mapsto -i\frac{z-2i}{z+2i}$$
for our conformal map from the upper half-plane to the unit disk. Composing the two conformal maps, we get $f = T \circ s \colon Q \to \mathbb{D}$, given by
$$f(z) = -i\frac{z^2-2i}{z^2+2i}.$$
(Note: That is the only map with the required properties; if $g \colon Q \to \mathbb{D}$ is conformal with $g(1+i) = 0$ and $g(0) = i$, then $g\circ f^{-1}$ is an automorphism of $\mathbb{D}$ that fixes $0$, hence a rotation, and also fixes $i$, hence the rotation must be the identity.)
Any Möbius transformation which maps the intersection point $z=1$ to $w=\infty$ and the real axis onto itself is a good start, because it maps the disc boundaries to vertical lines.
A possible choice is $T(z) = z/(z-1)$ which maps $1, 0, -1$ to $\infty, 0, 1/2$, respectively, and therefore the circle $\{ |z| = 1 \}$ to $\{ \operatorname{Re}(w) = 1/2 \}$ and the circle $\{ |z-1/2| = 1/2 \}$ to $\{ \operatorname{Re}(w) = 0 \}$, i.e. to the imaginary axis.
Now map one additional point to decide whether the image of your domain is the area between the vertical lines or the “outside.”
Best Answer
"Why finding a LFT which maps $(−i,i,\sqrt2−1)→(0,\infty,1)$, would be enough for the required mapping?"
An LFT is conformal in any case, so it is enough to verify the boundary of the image $I$.
The LFT maps a circle/line to a circle/line. So, the images of the circles $|z-1|=\sqrt2$ and $|z+1|=\sqrt2$ are circles or lines. The two corners $\pm i$ are mapped to $0$ and $\infty$, so those boundary curves pass through $\infty$, indicating that they are some straight lines passing through $0$.
The LFT preserves orientated angles; in particular, the right angle at $-i$ is mapped to a right angle at $0$; so the two boundary lines of $I$ are perpendicular. Therefore, $I$ is a right angled angle domain with vertex at $0$.
Finally, the boundary point $\sqrt2-1$ lies on the circle arc on right side of the first domain, and it is mapped to $1$, so $1$ lies on the lower leg of that angle domain. Tis fixes the direction for the right angle.
"Is the order of the points important? Why are the points $(0,\infty,1)$ taken in that order? I recall that I did find some transformations and the order was: $\to(1,0,\infty)$"
There are two questions here.
The first question is the orientation of the points. In case of such simple regions, a conformal map sends the oriented boundary of one region to the oriented boundary of the other region. So, the order of the 3 points around the boundary must be preserved. This accords to the argument principle: if you walk around $0$ along the boundary of the first region, your image should go $+1$ times around the image of $0$. So you cannot replace say $(0,\infty,1)$ by $(\infty,0,1)$.
The second question is about cyclic shifts of the points. In general, it is possible to replace $(0,\infty,1)$ by $(1,0,\infty)$ and obtain another conformal map, but the new map will not be an LFT, because the angles at the boundary are not preserved: the right angle at $-i$ is mapped to a straight angle at $1$.