For a solid surface moving in an air stream, a boundary layer forms in a region close to the surface, and the air velocity varies within that boundary layer, from zero relative to the surface to the free stream velocity outside the boundary layer, and the physical mechanism for that velocity gradient is air viscosity, or friction.
There is thus work done on the air in the boundary layer, causing heating and increased temperatures. For very high free stream velocities, the temperatures reached are high enough to dissociate the air molecules, producing plasma.
A very hot boundary layer conducts heat to the surface, increasing it's temperature enough to cause pyrolysis within a layer of the solid; i.e., a conversion of solid to gas. The latent heat of this conversion is the energy carried away from the surface; the resulting gas being incorporated into the boundary layer and carried away from the vehicle. This is a simple explanation of the ablation process.
I believe the questioner asks how the latent energy taken from the surface compares with the radiation of energy that will occur between the plasma in the boundary layer back to the surface. This "re-radiation" occurs because the new surface exposed when material is ablated will be at a lower temperature than the plasma. There is also re-radiation from the boundary layer to regions away from the craft. There is considerable complexity to a thorough understanding of the entire ablation process.
The simple answer is that we know this re-radiated energy back to the surface must be much less than the latent energy, otherwise ablation wouldn't work, and the surface would be destroyed. This also indicates that the re-radiation away from the craft must also be small, although greater than that going back to the surface, since the sink temperature of ambient is much less than the surface temperature. In addition, the shape factor for radiation away from the craft is probably greater.
With all that and in specific response to the questions:
How much heat does ablated matter in fact "carry away" as heated material?
The question, "How much," isn't defined, since it could indicate either a numerical answer or an answer describing relative magnitudes. Relatively, the ablated matter carries away the vast majority of the dissipated kinetic energy of the moving nose cone. If not, ablation would not work.
How much does it simply re-radiate back towards the plasma sheath?
Again, you're not clear on what you mean by "how much." In addition, you seem to be separating the ablated material from the plasma sheath. In reality, the ablated material contributes to the plasma sheath. But again, the amount of energy re-radiated back to the surface must be a minor part of the total energy carried away by the ablated material, otherwise ablation wouldn't work.
Can these be expressed as a fraction of the total energy dissipated stopping the capsule?
Yes, of course. The fraction due to ablation will be close to unity, since ablation is the major mechanism by which the nose cone slows down. A smaller fraction will be due to the energy radiated away from the nose cone, by all mechanisms, including radiation from the plasma envelope and radiation directly from the surface to which the plasma is more or less transparent. The particular numerical value will depend on the particular application, for instance, the nature of the ablated material, the geometry of the nose cone, the velocity in air, the altitude and density of the air, and other variables that render no one answer in a numerical sense.
Can they also be expressed in Joules/kg of ablated material?
Yes, of course. The kinetic energy of the moving nose cone can be expressed in Joules, and the mass of ablated material can be expressed in kg. The drop in velocity of the nose cone reflects the amount of kinetic energy dissipated, and the latter can also be expressed as Joules. This latter magnitude divided by the mass of ablated material in the process gives you the ratio you want. This ratio however does depend on the many variables I stated above, with a variation depending on the role of radiative energy from the craft.
