You can consider the union $C$ of the two circles (where the outer one is oriented one way and the interior the opposite way) as a cycle (i.e. formal sum of paths with boundary zero). Cauchy's theorem says that if the winding number is zero outside the domain in question (that is, outside the annulus), then the integral over $C$ of $\frac{f(\zeta)}{\zeta - z}$ is $2\pi f(z) n_C(z)$. The way the two circles are oriented is going to imply that $C $ has zero winding numbers about every point outside the annulus, so we get
$$f(z) = \frac{1}{2\pi i} \int_C \frac{f(\zeta)}{\zeta - z} d \zeta,$$
which is what you want.
I'm going to take $z_0=0$ for simplicity. For $\rho_1, \rho_2$ with $r_1 < \rho_1 < \rho_2 < r_2$, and for $z$ with $\rho_1 < |z| < \rho_2$, Cauchy's integral formula applied to the annulus $\{ \rho_1 \leq |w| \leq \rho_2\}$ gives us that
$$f(z)= \frac{1}{2\pi i} \int_{|w|=\rho_2} \frac{f(w)}{w-z} \ \mathrm{d}w - \frac{1}{2 \pi i} \int_{|w|=\rho_1} \frac{f(w)}{w-z} \ \mathrm{d}w.$$
In the first integral we can write
$$\frac{f(w)}{w-z} = \sum_{n=0}^\infty \frac{z^nf(w)}{w^{n+1}},$$
and for each $z$ with $|z| < \rho_2$ this sum converges aboslutely uniformly on the circle $\{|w|=\rho_2\}$.
[Added in response to comment: there exists a positive real number $M$ such that $|f(w)| \leq M$ on $\{|w|=\rho_2\}$. Then for all $w, z$ with $|w|=\rho_2$ and $|z|<\rho_2$, and all $n \geq 0$, we have
$$\Bigg|\frac{z^nf(w)}{w^{n+1}}\Bigg| \leq \frac{M}{\rho_2} \Bigg(\frac{|z|}{\rho_2}\Bigg)^n.$$
The right-hand side converges (it's a geometric series with ratio $|z|/\rho_2 <1$) so the claimed sum converges absolutely uniformly on the circle by the Weierstrass M-test.]
So the first integral is equal to
$$\sum_{n=0}^\infty z^n \frac{1}{2\pi i} \int_{|w|=\rho_2} \frac{f(w)}{w^{n+1}} \ \mathrm{d}w,$$
and this converges and is valid on the whole of $\{|z| < \rho_2\}$. Letting $\rho_2 \rightarrow r_2$, we see that the positive part of the Laurent series converges on $\{|z| < r_2\}$.
For the negative part, do a similar expansion with $w^n/z^{n+1}$.
Best Answer
I assume you want the Laurent series expansion for this function at $0$ (since this is the point those annuli are centered at). The standard way to do this is not to use the integral formula at all--- just the uniqueness of the Laurent series expansion (which I guess is often proved using the integral formula). You do some algebra to find series representations of this function in integer powers of $z$ that converge in the given annuli. By uniqueness these series have to be the Laurent series.
From partial fractions $$ \frac{1}{(z - 1)(z - 2)} = -\frac{1}{z-1} + \frac{1}{z-2} $$ so on either annulus it is enough to find the Laurent series expansions of the functions $\frac{1}{z-1}$ and $\frac{1}{z-2}$ and then combine them term by term.
If $|z| > 2$ then note that $$ \frac{1}{z - 2} = \frac{1}{z} \frac{1}{1 - 2z^{-1}} $$ and the fact that $|z| > 2$ implies that $|2 z^{-1}| < 1$. If you recall the geometric series formula, $$ \frac{1}{1-u} = 1 + u + u^2 + \cdots, \qquad |u| < 1, $$ then putting in $u = 2 z^{-1}$ you see that $$ \frac{1}{z - 2} = \frac{1}{z} (1 + (2z^{-1}) + (2z^{-1})^2 + \cdots) $$ holds for all $|z| > 2$. If you just expand this out you get the Laurent series for $\frac{1}{z - 2}$ in this region. Of course it is much cleaner if you use sigma notation to write it out.
On the annulus $1 < |z| < 2$ of course the above method does not work. But we also have $$ \frac{1}{z - 2} = -\frac{1}{2} \frac{1}{(1 - \frac{z}{2})} $$ and if $1 < |z| < 2$ then $|\frac{z}{2}| < 1$ so the geometric series formula can be used again in a slightly different way here.
The Laurent series expansion of $\frac{1}{z - 1}$ is easier since $\frac{1}{z-1} = \frac{1}{z} \frac{1}{1 - z^{-1}}$ and $|z^{-1}| < 1$ holds for all $z$ in either annulus. So the Laurent expansion of this function is the same in both regions.
The ideas used here generalizes to finding the Laurent series other rational functions on other annuli of course.