I don't know the reference for the answer, so let me propose the asnwer in the case
when S has negative Euler characterisitc, orientable and CONNECTED. At least you can compare with your own, answer, hope that it is correct. Let p(s) denote the number of punctures.


1) chi(S')/chi(S)=d with $d$ positive integer

2) $p(S)\le p(S') \le p(S)d$. 

3) $p(S)d-p(S')$ should be even. 

Moreover, in the case Genus(S)=0 you have an addtional condition
  
$p(S)d-p(S')>d-2$.

This condition comes if we want S' to be connected. If S' is alowd to be disconnected, this condition desapears.

I hope that the answer is correct. All the conditions seem to be necessary (and apearently sufficient). Condition 3 comes from the fact, that the permutation corresponding to going around all punctures on S is a commutator, so it should be even. I think, the proof should not be too involved. Just using some standard facts about permutation groups. I am not 100% sure if you need to include it...