How does the polarization of the carbonate ion make the thermal decomposition of CaCO3 more likely?

I think I see the confusion. You may not be seeing the intramolecular effects. The decomposition reaction is:

#"CaCO"_3(s) stackrel(Delta" ")(->) "CaO"(s) + "CO"_2(g)#

The effect of interest is the polarization towards the #"Ca"^(2+)# cation from the #"CO"_3^(2-)# anion, causing intramolecular bond weakening.

Yes, there is an ion-pairing attraction going on, but that is not going against the ability of the #"CaCO"_3# to decompose when you heat the compound (it is a #"Ca"^(2+)-"O"^(2-)# ion-pair interaction that will later hold the #"CaO"(s)# formula unit together, anyways). However, what you should also be looking at is how the polarization facilitates decomposition.

Since the calcium cation is highly-positively-charged, and is somewhat small, one might call it a "hard acid" (from Hard-Soft Acid/Base Theory), because it can capably concentrate negative charge density towards itself, and we call that great polarizing ability. We can also say that #"Ca"^(2+)# is very electropositive.

Thus, it pulls electron density from the #"CO"_3^(2-)#, specifically within the #"C"-"O"# bond (since the individual dipoles [that all cancel out] within #"CO"_3^(2-)# lie along the #"C"-"O"# bond, making it have the highest vector alignment). This unbalances the electron distribution and makes the #"C"-"O"# bond more ionic in character. That weakens the #"C"-"O"# bond, and facilitates breaking the bond.

That bond will be broken upon heating in order to perform the decomposition reaction.