While the heatspreader effect won't be fully quantified until folks can do back to back tests (with the same chip) using different heatsinks, we can make a bit of a guess.
AMD have used indium solder onto gold plate onto the main heatspreader which is primarily copper with a Nickel plate on top.
So,
- The solder has a thermal conductivity of 80 W/mK.
- The gold has a thermal conductivity of around 300 W/mK
- The copper has a thermal conductivity of around 400 W/mK
- The Nickel has a thermal conductivity of around 90 W/mK
Good thermal paste has a thermal conductivity of around 8.5 W/mK.
Intel's TIM has a thermal conductivity of around 6 W/mK.
http://www.techarp.com/articles/delidded-threadripper-secrets-revealed/
Now, lets assume that the heat will pass through the gold/copper/nickel mix. Because the copper is most conductive, we will further conservatively assume it passes only through the copper - which we'll assumed is 1.25 mm thick in an IHS of overall thickness 1.5mm.
If we further assume only 70% of the die has normal conductance through the heatspreader to heatsink then the average distance the heat must travel is approx ~.0042m (based on full diameter of 51.9mm to enclose all 4 dies and thus 43.5mm dia to cover 70% of die area).
Take the conductive area as the circumference of 43.5mm cylinder with thickness 1.5mm and for copper you get a thermal resistance of 0.16 K/W.
Then, with 30% of 180W passing through this, the outside corners of the CPU dies will see a temperature rise of between 10 and 17 Kelvin (worst case scenario since we haven't added in the nickel or gold or indium solder).
Given that reviews are recording TR temperatures ~65degC at stock and ~80-85degC on overclock, and compare that to the Intel Core i9s... I don't think the TR heatsink/heatspreader incompatibilities are much of an issue.
[Saturation should be a non-factor as the heatsink will be continually extracting heat from the system and the overall thermal gradients should be large enough to render the localised effects rather minor.]