To explore the contributions of electronic (ion-electron scattering) and nuclear (ion-target atom) stopping, you can look at SRIM, available for download free from srim.org. This program was create by Ziegler, Ziegler, and Biersack ages ago and has been update by them since (yes, the same Ziegler as on a label on your graph above). Note that SRIM does not deal with the radiative effects at very high particle energy - it is/was mainly concerned with modeling ion implantation effects, not relativistic high energy particle physics.
Calculating the stopping power/ranges for protons in water, it is readily seen that the electronic stopping dominates nuclear interactions along almost all of the proton path length (again, I'm ignoring radiative contributions). A few examples are given here (stopping units of MeV/(mg/cm2)):
Proton energy dE/dx (elec) dE/dx (nucl) Range +/- straggle (um)
10 MeV 4.657E-2 2.700E-5 1200 +/- 54
1 MeV 2.402E-1 2.090E-4 26.55 +/- 1.16
100 keV 8.252E-1 1.476E-3 1.43 +/- 0.11
10 keV 4.283E-1 8.66E-3 0.26 +/- 0.07
1 keV 1.447E-1 3.364E-2 0.03 +/- 0.02
100 eV 4.577E-2 6.354E-2 0.004 +/- 0.004
So, for protons into water, the peak of the overall stopping is near 100keV, and almost all of that stopping is electronic. Only at very low energies does the nuclear stopping finally exceed the electronic. This is why protons have a pretty sharp end of range - the electronic stopping is a very smooth stopping as the proton scatters off of lots of electrons, losing very little energy each time. The nuclear stopping is a more violent event (larger momentum transfer), but really only happens at the end of range. So, the proton transfers energy to target atoms only at the end of range. Energy is being continually lost to electron scattering along the whole trajectory (in varying amounts depending on the proton energy at that point).
That last bit is important - you can get the energy loss at a given energy, but as the proton loses energy, you have to re-evaluate what the energy loss is at that new energy. If you are above the energy loss peak, a little more proton energy gives you a much longer range until it finally slows closer to the peak energy loss. (Note that a 10MeV proton has a range of 1.2 millimeters, while a 1 MeV proton goes only 26 microns, and a 100keV proton about 1.5 microns.)
Best Answer
"STOPPING POWER refers to the inelastic energy losses by an electron moving through a medium ... which represents the kinetic energy loss by the electron per unit path length"
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"LINEAR ENERGY TRANSFER of charged particles in a medium is the quotient dE/dl, where dE is the average energy locally imparted to the medium"
So the linear energy transfer is the amount of energy (per length) the medium receives from the particle, while the stopping power is the amount of energy (per length) the particle loses. The difference between these two is the amount of energy (per length) that is lost to the rest of the environment. These are called radiative losses.
Radiative losses occur when the electron is accelerated-- when this happens, it releases some energy into the electromagnetic field in the form of a wave. It would happen even in the absence of a medium if the electron were accelerated in some other fashion (say, if it's going around a ring in a particle accelerator). The energy lost (per time) from a uniformly accelerating charge to the electromagnetic field is given by the Larmor formula, $$P = \frac{q^2 a^2}{6 \pi \epsilon_0 c^3}$$