Since you are interpolating the value of specific internal energy "u", you are observing this difference.
You can compute specific enthalpy "h" from the specific internal energy "u", pressure "p", and specific volume "v":
>> u_sat_v = interp1( r134aPropertyTables.p, r134aPropertyTables.vapor.u_sat, p)
>> v_sat_v = interp2( r134aPropertyTables.p, r134aPropertyTables.vapor.unorm, r134aPropertyTables.vapor.v, p, 1)
>> h_sat_v = u_sat_v + p*1e3*v_sat_v
To get "v_sat_v", I interpolated the table at unorm = 1, where "unorm" is what we call the normalized internal energy, because that corresponds to the vapor saturation line.
For liquid, specific volume is very small, so specific enthalpy and specific internal energy are very close in value.
In general, please be careful about comparing values of enthalpy, internal energy, and entropy from different sources because they are state properties, which means they can have an arbitrary reference value.