What is the explanation for this stereochemical product?
No, you’re overthinking this. It’s just sterics.

The iodine will form a iodonium ion (three-membered ring with positive charge on iodine) where the double bond was. Since iodine is huge, it will come from the less sterically hindered side which is from behind the monitor plain.

The carboxylic acid then attacks the iodonium rom above closing a five-membered lactone ring (kinetically favoured against a six-membered one). The side chain carbon retains iodine pointing backwards (you can twist along the $\ce{C-C}$ bond to push the carbon into the monitor plain and iodine backwards) while the lactone oxygen coming from above means that the methyl substituent on the ring must be pointing behind the plain, again.

Also, stereochemistry on the hydroxide should be retained since it does not take part in the reaction.


Scheme 1: A not-well drawn mechanism for the reaction.