Create a transformation from $(z_1,z_2,z_3)=(0,i,\infty)$ to $(w_1,w_2,w_3)=(2,1-i,0)$, the transformation should be $w(z)=\frac{2}{z+1}$
My calculations: $$\frac{w-2}{w-1+i}=k\frac{z-0}{z-i}$$ By substituting $z=\infty$ and $w=0$ I get $k=\frac{2}{1-i}$.
Multiplication of the left denominator to the right part gets me: $$w-2=w\frac{2z}{(1-i)(z-i)}-\frac{(1-i)\cdot 2z}{(1-i)(z-i)}$$ $$w=\frac{2+\frac{2z}{z-i}}{1-\frac{2z}{(1-i)(z-i)}}=\frac{2(z-i)+2z}{(z-i)-\frac{2z}{1-i}}$$
This feels correct but from here on my calculations gets me nowhere...
At which point do I make the wrong calculations?
Note that $k = \frac{2}{i-1} = 1+i$. This simplifies the calculations slightly. I continue where you, in my opinion, went off in the wrong direction. You should've multiplied away all the denominators, then gathered all $w$ terms on one side, like so: $$ \frac{w-2}{w-1+i}=(1+i)\frac{z}{z-i}\\ (w-2)(z-i) = z(1+i)(w-1+i)\\ (z-i)w - 2z + 2i = (1+i)zw - 2z\\ (-iz-i)w = -2i\\ (-iz-i)w\cdot i = -2i\cdot i\\ (1+z)w = 2\\ w = \frac{2}{1+z} $$
I would solve it by beginning with the function $w(z)=\frac{az+b}{cz+d}$ and find $a,b,c,d$ that solve the equations $$w(0)=2\\w(i)=1-i\\w(\infty)=0$$ because I think it's easier to see why this actually solves the problem. But that's just personal taste. The downside is, of course, that if we don't have a lot of $0$ and $\infty$ in our points, this becomes a bit more calculation heavy. Luckily, there are plenty of $0$ and $\infty$ to go around this time. The third equation tells you $a=0$. The first equation tells you that $b=2d$. Inserting this into the second equation gives $$ \frac{2d}{ci+d}=1-i $$ which becomes $c=d$. Finally, setting $c=1$ gives $w(z)=\frac{2}{z+1}$.