In Carruthers' Modern Methods of Organic Synthesis, the following aldol reactions of two (Z)-enolates with benzaldehyde are shown:
The two transition states shown seem to be different, and lead to different stereochemistry in the product (the first product has both Me and OH pointing away from the plane, whereas the second has both pointing out of the plane).
Is this significant, and if it is, how can we predict which TS is used for which problem specifically?
Ignoring the difference between the SPh and Et groups, the two products shown are enantiomeric, not diastereomeric. Both of these reactions are not enantioselective in any manner, so both enantiomers will be formed in equal amounts, via both transition states. Presumably, in the book, only one enantiomer is shown to save space and to avoid redundancy.
The only thing that needs to be illustrated here is that the phenyl group is equatorial in the lower-energy transition state. That leads to the diastereoselective formation of the syn aldol product (but no enantioselectivity between the two syn products).
So, there is no difference at all between the two models you have used. Both are the same thing and are equally applicable. If you had a chiral auxiliary on boron or something of that sort, the story might be different. But that isn't the case here.
