"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 $1$.