You can see the bremsstrahlung at the left of the graph; there are two peaks, labeled $K\alpha$ and $K\beta$, which are due to inner orbital electrons being ejected from the K-shell of the atoms by the x-ray source. The ejected electrons are replaced electrons from other shells, and the characteristic energy of the two peaks is due to the energetic photons released when electrons "fall" into the vacancies.

The other peaks, further to the right, at higher angles, are the x-ray diffraction peaks.

You can see the bremsstrahlung at the left of the graph; there are two peaks, labeled $K_\alpha$ and $K_\beta$, which are due to inner orbital electrons being ejected from the K-shell of the atoms by the x-ray source. The ejected electrons are replaced electrons from other shells, and the characteristic energy of the two peaks is due to the energetic photons released when electrons "fall" into the vacancies.

The other peaks, further to the right, at higher angles, are the x-ray diffraction peaks.

Usually the $K_\alpha$ and $K_\beta$ are the source x-rays, and provide the monochromatic beam which illuminates the crystal or powder; in a typical device these are blocked with a beam stop; otherwise they are too bright, and don't contribute anything. In this case it appears that they may have been generated inside of your material, but that depends on the instrument, the target material, and the energy of the source x-rays.

In the case of NaCl the characteristic x-rays are of fairly low energy, and there will be characteristic peaks for both atomic types, Na and Cl.

The source of the x-ray doesn't matter; everything will contribute to the Bragg diffraction. The beam with the highest intensity will dominate, which should be the source x-ray. If the diffracting beam is not monochromatic, due to the multiple contributions, the diffraction peaks will be broadened, and may even split. You can calculate this base on the energies used.