I am considering the hydrogenation of 2-butanone with respect to its Re and Si faces.
So if we look at 2-butanone as I've drawn below from above we are looking at its rectus face.
And from below we are looking at its sinister face.
Now, time to consider hydrogenation.
We'll keep the wedge and dash groups as they are and the carbon and oxygen both on lines. In other words we keep two points as is in the originally trigonal planar 2-butanone and when the hydrogen inserts itself, the other two points (a hydrogen and an alcohol group) pop up out the plane to form the classic tetrahedral shape of an $\ce{sp3}$ carbon atom.
So it appears that hydrogenation from the sinister face adds an hydrogen to the central carbon and forms an alcohol, and the alcohol is above the added hydrogen spatially, and this gives us the R isomer. Hydrogenation from the rectal face gives the opposite S isomer.
How's the reasoning? Am I right? I am fairly sure. But not completely sure.