I wish to find the Laurent series for the function
$$f(z) = \frac{1}{z(z-1)}$$
on the annulus $2 < |z+2| < 3$.
From the bounds of the domain I deduce that we would like something of the form $$\frac{|z+2|}{2} > 1, \quad \frac{|z+2|}{3} < 1$$ so that we can use geometric series about the point $z = -2$. Furthermore, using partial fraction decomposition I have that $$f(z) = \frac{1}{z-1} - \frac{1}{z}.$$
My question now is before manipulating the above any further, how do I know which of the above two fractions to expand as a Taylor series and which as a Laurent series? The texts I have referred to all gloss over this heuristically, which makes me believe I am missing obvious.
You wrote "about the point $z=2$", but I think you mean about the point $z=-2.$ That is your home base. Let $f_1(z)=1/(z-1)$. This $f_1(z)$ is analytic in a disc (centred) at $-2$, of which your annulus is a subset.
Let $f_2(z)=1/z$. This $f_2(z)$ is NOT analytic in any disc at $-2$, of which your annulus is a subset. So this is the one for which you need to use Laurent's Theorem.
(Of course $f_2$ is analytic in small discs at $-2$, but these don't help you.)
You should, if you haven't yet, sketch the annulus and mark the two isolated singular points of $f(z)$.