Power resistors equivalent thermal model: Interpretation of datasheet & 0°C terminal of power source

One or more of your assumptions is not correct.

You should have \$\theta_{ja} = \theta_{jc} + \theta_{ca}\$, giving \$\theta_{ca}\$ of 30.76 C/W for IC1 and 21.76 C/W for IC2.

With those values, IC1 should have a case temperature of

23 C + (1.2 W) (30.76 C/W) = 59.9 C,

which is pretty close to what you measured; but IC2 should have

23 C + (0.7 W) (38.23 C/W) = 38.23 C,

which is way off the measurement.

A likely cause are if the datasheet's \$\theta_{ja}\$ value makes some assumption about board layout that is not met in your application. If your board doesn't provide sufficient copper area attached to the chip to spread heat away, \$\theta_{ja}\$ will increase substantially. It's also possible that the datasheet value assumes some forced air cooling over the chip.

Some of this is partially answered in the comments to the question, but here's the complete (IMHO) answer:

\$R_{ja}\$ is the thermal resistance from the diode junction to ambient air assuming (1) no air movement (or perhaps natural conduction), and (2) there's no significant thermal coupling between the diode's leads and the ambient environment. It says, in effect, "if you take no special effort to couple (or insulate) this diode to (or from) the environment, this is what your thermal resistance is".

\$R_{jl}\$ is the thermal resistance from the diode junction to the leads -- or, if the diode is fully specified, to some point on the leads (i.e., 10mm from the case, or 5mm -- there's no standard, read the data sheet carefully). Using this you can arrange your application (somewhat) to suit the diode, by arranging to couple the leads to the ambient environment. Usually this is done by making extra-big traces on the PC board, or extra-big pads, to conduct heat from the diode leads to the surrounding PC board.