Will a Grignard reagent react with an epoxide or episulfide first?
There are quite a few reasons why the episulfide might open before the epoxide:


Sulphur is less electronegative than oxygen. Therefore, by Bent's rule , more atomic s-character will be directed by the ring towards sulphur in the episulfide as compared to the oxygen in the epoxide. An increased s-character will lead to a bigger increase in the $\ce{C-S-C}$ bond angle as compared to the $\ce{C-O-C}$ angle, and thereby increasing the angle strain in the episulfide relative to the epoxide
Sulphur will also tend to be a better leaving group, as basicity decreases down the group(in this case, group 16)
The carbon-sulphur linkage is weaker than the carbon-oxygen one


So, the mechanism should look like this:



Some points to note:


Sulphur compounds also tend to be pretty good nucleophiles due to their softness as per HSAB as elaborated here. Therefore, the $\ce{RS-}$ formed after ring opening will again attack on the epoxide to open the latter. 
This will also lead to the formation of an  ethereal linkage in the final product, which will be quite favorable due to the high stability of $\ce{C-O}$ ether bonds