Enantioselectivity of Noyori Asymmetric Hydrogenation
I think this is easy to answer than you think. Let's look at the stereochemistry of compounds in the question, 1g and 1h (I also put the common product, 1a, for convenience):



The stereospecific reduction is on carbon-3 ($\ce{C}$3) in all cases. Accordingly (see diagram above), if you apply Cahn–Ingold–Prelog (CIP) sequence rules for each, you'd find (3R)- for 1a (methyl group is priority 3 based on $\ce{CHH}$ vs $\ce{HHH}$), (3S)- for 1g (isopropyl group is priority 2 based on $\ce{CHH}$ vs $\ce{CCH}$), and  (3S)- for 1h (phenyl group is priority 2 based on $\ce{CHH}$ vs $\ce{CCC}$). All these assignments are $\ce{C}$2 versus $\ce{C}$4 since priory 1 for all three groups on $\ce{C}$3 is $\ce{OH}$. 

Thus, we can conclude that the change of stereochemistry is mainly due to the priority rules (not due to the change in stereochemistry on transition states or intermediates). This is also supported by the example given in the same reference (Ref.1) in the box in depicted diagram. Even when bulkiness at $\ce{C}$2 was increased by adding a methyl group, the reaction (with R-BINAP) has undergone same manner to give (R)-configuration to $\ce{C}$3 center.

References:


Ryoji Noyori, Takeshi Ohkuma, Masato Kitamura, Hidemasa Takaya, Noboru Sayo, Hidenori Kumobayashi, Susumu Akutagawa, "Asymmetric hydrogenation of $\beta$-keto carboxylic esters. A practical, purely chemical access to $\beta$-hydroxy esters in high enantiomeric purity," J. Am. Chem. Soc. 1987, 109(19), 5856-5858 (https://doi.org/10.1021/ja00253a051).