Consider the integral $$ \oint dx\, \frac{g(x)}{x - x_{0}} $$ where $x_{0} \in \mathbb{C}$ is the position of a pole for the integrand that we assume to be integrated along a closed path encircling this pole and all other poles of $g(x)$.
The residue theorem would suggest that the result of this integral is $$\pm 2\pi i \Big[ g(x_{0}) + \sum_{\textrm{poles}} \frac{\textrm{res}(g)(\hat{x})}{\hat{x} - x_{0}}\Big]\,$$ depending on the orientation of the contour about the pole. In particular, if $g(x) \sim \ldots + \frac{B}{x} + C + \ldots$ about $x = 0$ then we find a contribution $-B/x_{0}$ from the second term as $x\rightarrow 0$ (which we have said will be inside the contour).
Now suppose $g(x) = xf(x)$. Then $g(x_{0}) = x_{0}f(x_{0})$ and $\textrm{res}(g)(\hat{x}) = \hat{x}\,\textrm{res}(f)$ for $\hat{x} \neq 0$. However, if I evaluate this integral as follows I get a different result:
Write $\frac{xf(x)}{x-x_{0}} = f(x) + x_{0}\frac{f(x)}{x-x_{0}}$. The first term gives some contributions from the same poles of $g$ as above, plus a term proportional to $C$. The second term cancels the $C$ part, provides $x_{0}f(x_{0})$ and turns those other pole contributions into the same ones as above -- except: there is no contribution from the pole in $g$ as $x \rightarrow 0$!!
What is the problem here - is it that I can't apply linearity of the integrand when the contour lives in two different spaces (i.e. $\mathbb{C}$ and $\mathbb{C} \diagdown \{0\}$)?
Assume $g(x)$ has Laurent expansion around $0$ given by $\sum_{n=-\infty}^\infty a_n x^n$ and $a_{-1} \neq 0$. Then $f$ has Laurent expansion around $0$ given by $\sum_{n=-\infty}^\infty a_{n+1} x^n$. As you can see, $\operatorname{Res}(f,x,0)$ may or may not be $0$ in this situation; it depends on whether $a_0=0$, and we didn't assume anything about that.
However, $\operatorname{Res}(x_0 f(x)/(x-x_0),x,0)$ is not just $-\operatorname{Res}(f(x),x,0)$. Because if you expand the contribution of the denominator around $0$ then you get $-f(x)/(1-x/x_0)=-f(x) \sum_{n=0}^\infty (x/x_0)^n=-\left ( \sum_{n=-\infty}^\infty a_{n+1} x^n \right ) \left ( \sum_{m=0}^\infty (x/x_0)^m \right )$. So the desired residue is obtained by summing over $n<0$ with $m=-1-n$; thus you get $-\left ( \sum_{n=-\infty}^{-1} a_{n+1} x_0^{n+1} \right )$. The term $n=-1$ picks up the residue from $f$ itself, but there are more terms here if $f$ has a higher order pole at $0$. There is at least one of those, because you assumed $a_{-1} \neq 0$.
Basically your error is applying the residue operator to $g$ alone instead of the entire quantity $g/(x-x_0)$. For a simple pole in the numerator this is misleading but gets the right answer anyway. Whereas if you have a higher order pole in the numerator then this is just incorrect.