I am calculating Laurent series of $$ f(z)=\frac{1}{(z-1)(z-2)} $$ which converges when $|z-1|>1$.
I started as $$ \frac{1}{(z-1)(z-2)}=\frac{1}{z-1}\cdot\frac{1}{(z-1)-1}=\frac{1}{(z-1)^2-(z-1)}. $$ Is this even the right way to transform function to some kind of geometric series, which can be easily expandet as Taylor series? How to calculate this Laurent seires?
You are on the right track. Now note that
$$\frac1{(z-1)(z-2)}=\left(\frac1{(z-1)^2}\right)\left(\frac1{1-\frac1{z-1}}\right)$$
Now, expand the second term in a Taylor series of powers of $\frac1{z-1}$ (It's a simple geometric series.).
Can you finish now?
You were doing it well:
$\frac{1}{(z-1)-1}=\frac{-1}{1-(z-1)}=\sum_{n=-\infty}^{-1}{(z-1)^n}=\frac{1}{z-1}+\frac{1}{(z-1)^2}+...$
Finally just multiply the series above by $\frac{1}{z-1}$
$$f(z) = \frac{1}{(z-1)(z-2)}$$
We write $f(z)$ in its partial fraction expansion, and then we expand $f(z)$ in powers of $\frac{1}{z-1}$:
$$\begin{aligned} f(z) &= \frac{1}{z-2}-\frac{1}{z-1} \\ &= \frac{1}{(z-1)-1} - \frac{1}{z-1} \\ &= \frac{1}{z-1} \frac{1}{1-\frac{1}{z-1}}- \frac{1}{z-1}\\ &= \frac{1}{z-1} \left( \frac{1}{z-1} + \frac{1}{(z-1)^2} + \cdots \right) \\ &= \frac{1}{(z-1)^2} + \frac{1}{(z-1)^3} + \cdots \end{aligned}$$ This is the Laurent series for $f(z)$ in the region $|z-1|>1.$
