Why is it that we can't apply the residue theorem $\int_\gamma f(z)dz = 2\pi i\sum_nres(f,c_n).ind(\gamma,c_n)$ with $f$ non defined on several disks of center $c_n$, and that we can only do it with $f$ non defined on the points $c_n$ themselves?
I mean, if $\int_\gamma f(z)dz = 2\pi i.res(f,c).ind(\gamma,c)$ is correct on an open ring of center $c$, can't we just see the integral on a domain missing several disks as the addition of integrals on rings homotopic to some sub-domains (covering the full domain altogether) around each of those disks?
Conceptual answer
Think about how we prove the Residue Theorem. Typical proofs work by splitting $\gamma$ into a sum of curves around each singularity, then applying results like Cauchy's Theorem and Cauchy Integral Formula to individual terms in a Laurent series. The proof of Cauchy Integral Formula basically comes down to shrinking the curve $\gamma$ to a smaller and smaller neighborhood of the singularity.
If your function is only defined via a Laurent series that converges on some annulus that isn't a punctured disc, then these proofs will fail at the "curve shrinking" step.
Alternatively, many (most) would probably say that the proofs fall apart before that step: it doesn't really make sense to define "residues" or decide which type of singularity we're looking at if all we have is a Laurent series that converges on a non-punctured-disc annulus. To see why we run into problems if we try to define and use those concepts on larger annuli, here's an.....
Explicit example
Check out this MSE answer. It gives an explicit example using two different Laurent series for $f(z) = \frac{1}{z(z-1)^2}$. If you (incorrectly) reason from the Laurent series that converges on $|z-1|>1$, then you'll end up concluding the residue at $z=1$ is 0. If you (correctly) use the Laurent series that converges on $0 < |z-1| < 1$, then you'll find that the true residue is -1. These don't match, so using the bigger annulus would get you the wrong answer.