Assume the function $f(z)$ has a simple pole at $z_{0}$.
There is a theorem that states that if $C_{r}$ is an arc of the circle $|z-z_{0}| = r$ of angle $\alpha$, then $$\lim_{r \to 0} \int_{C_{r}} f(z) \, dz = i \alpha \, \text{Res}[f,z_{0}].$$
But what if $z_{0}$ is a pole of higher order? Can we say anything definitive about $ \lim_{r \to 0} \int_{C_{r}} f(z) \, dz $?
On
I think you are going to have problems with divergences. In your example above with the double pole, you may parametrize the integral as $z=z_0+\epsilon\, e^{i \phi}$, for $\phi \in [\phi_0,\phi_0+\alpha]$. Then the integral over the arc is
$$i \epsilon \int_{\phi_0}^{\phi_0+\alpha} d\phi \, e^{i \phi}\, f(z_0+\epsilon\, e^{i \phi}) \sim \frac{i}{\epsilon} \int_{\phi_0}^{\phi_0+\alpha} d\phi \, e^{-i \phi}$$
Note that, unlike the case where you integrate above a closed curve, the integral does not vanish in general. When the integral does not vanish, the integral about a double pole (or greater, say, $n$) blows up as $\epsilon \to 0$ unless you go around a fully closed curve. (There are cases, of course, where the integral does vanish, i.e., when $n \alpha = k 2 \pi$).
For $n \geqslant 2$, the function $\dfrac{1}{(z-z_0)^n}$ has a primitive $\dfrac{(-1)}{(n-1)(z-z_0)^{n-1}}$, so if the arc is $z_0 + re^{it}$ for $\varphi \leqslant t \leqslant \vartheta$,
$$\int_{C_r} \frac{dz}{(z-z_0)^n} = \frac{1}{(n-1)r^{n-1}}\left(e^{-i(n-1)\varphi} - e^{-i(n-1)\vartheta}\right).$$
In general, that is unbounded for $r \to 0$, but for a given $n$, there are choices of $\varphi$ and $\vartheta$ that make the integral vanish.
If the principal part of $f$ has more than one term of order $< -1$, the choices for the difference between the two angles that make the integral vanish are even more restricted.