Show that the residue is $c_{-1}=-frac{q''(z_0)}{(q'(z_0))^3}$.

Why can we do that?

It can be done with a simple substitution: set $t=z-z_0$. Then $z=z_0iff t=0$, and the function $f(z)$ becomes the function $g(t)=dfrac 1{bigl(q(z_0+t)bigr)^2}$

Is this a Laurent series? What does the $O$ mean?

It is not a Laurent series, since it is not a series. It is just the Taylor's expansion of the Taylor series of $q(z)$ at order$2$, the remainder $r(z)$ being $O(z^3)$ in Bachmann notation (asymptotic analysis), which means that $dfrac{r(z)}{z^3}$ is bounded when $zto 0$.

How did we get the $:q'(0) q''(0)z^3+O(z^4),$?

Simply calculating the polynomial part of $bigl(q(z)bigr)^2$ and applying the rules of calculation with $O$, i.e. truncating everything with degree $ge 3$ (it becomes a part of $O(z^3)$)

How does this get us the desired result?

After factoring out $(q'(0)^2)z^2$ in the denominator, they're left with $$frac{1}{1+underbrace{frac{q''(0)}{q'(0)}z+O(z^2)}_{u}}$$ which they expand at order $1$ with the usual formula, truncating it again, at the same order as the denominator.

So they obtain begin{align} frac{1}{q(z)^2}&=frac{1}{(q'(0))^2,z^2}, frac{1}{1+cfrac{q"(0)}{q'(0)}z+O(z^2)}\& =frac{1}{(q'(0))^2 ,z^2}biggl(1-frac{q''(0)}{q'(0)}z+O(z^2)biggr) \[1ex] &=frac{1}{(q'(0)^2)},frac1{z^2}-frac{q''(0)}{(q'(0))^3},frac 1z +O(1). end{align}

Is this clearer?