Problem #1 is that if you interchange the methyl and hydrogen groups, you get diastereomers, not enantiomers. But this is a relatively minor point. I assume your question is about the diastereoselectivity, not enantioselectivity (as ron rightly pointed out there is no enantioselectivity in this reaction).
The analysis you have drawn is somewhat similar to a Fukui frontier molecular orbital analysis. In that picture, for a reaction to be thermally allowed, you need to check for constructive bonding overlap between the HOMO and the LUMO, within geometrical reason. This is achieved by connecting lobes of "like" shading: i.e. shaded lobe with shaded lobe, and unshaded with unshaded.
You don't necessarily have to draw it stepwise (although I gather it's your own way of doing it). You can simply draw both interactions in the starting material, like this. The bottom interaction is not an issue. The top interaction, however, is an issue.
Problem #2 is that you tried to connect the large unshaded lobe of the σ* orbital with the white lobe of the π orbital. That is not geometrically reasonable: if you connect these two lobes, then you need to somehow end up with the new σ bond being formed underneath the ring. In other words, you would have to have an antarafacial migration of the alkyl group.
Problem #3 is that you have concluded that the reaction must be suprafacial based on electron counting. Note that in this context, the reaction being suprafacial means that the migrating alkyl group stays on the same face of the ring. In fact, the reaction must occur in a suprafacial manner simply because of geometrical constraints: there's simply no way that the alkyl group is going to migrate to the opposite side. Technically it could, but you would have to break so many bonds in the transition state that it would not be possible.
[Note that this has a different meaning from the suprafacial or antarafacial components in a Woodward–Hoffmann analysis.]
So what is the solution? Well, because the reaction is suprafacial, you need to use the shaded lobe of the π HOMO, which is on the top face of the ring – the same face as the migrating alkyl group. So then for constructive bonding you need to also use the shaded lobe of the σ* LUMO.
Now one might argue that just like how the unshaded lobe of the π HOMO was geometrically inaccessible, so should the shaded lobe of the σ* LUMO. But we already know that the C–C σ bond is going to break in the course of the reaction, so you're going to go through a pathway whereby the bond behind it (with an arrow pointing towards it) can rotate such that the shaded lobe of the σ* can overlap with the shaded lobe of the π* orbital. In fact, that's not too different from what you drew. The transition state is pretty similar to the intermediate in your stepwise reaction – it just has a few more dotted lines.
If you now "join" the shaded lobes then you can see how the correct stereochemistry arises.
In general my preferred way of analysing pericyclic reactions is to use Woodward and Hoffmann's original rules. Here, there is no shading of lobes: you connect the lobes within geometrical reason, then count the number of $4q+2$ suprafacial components, and the number of $4r$ antarafacial components. If the total is odd, the process is thermally allowed; if it is even, the process is thermally forbidden.
I am not going to explain this approach in detail - it is too long for here and it is adequately covered in textbooks which you should have access to - but I will upload the two diagrams I drew: