I think your understanding of the Breit frame is not correct. From https://en.wikipedia.org/wiki/Breit_frame, the Breit frame corresponds to the frame in which the momentum of a scattered particle is the opposite of its momentum before scattering, it doesn't impose $q^2 = -\mathbf{q}^2$. $q^2$ is equal to $ -\mathbf{q}^2$ if the mass of the nucleus is infinite and we actually have $q^2 = \omega^2-\mathbf{q}^2$ where $\omega$ is the energy of the recoiled nucleus.

1. A photon is an elementary particle so there is no form factor.

2. & 3. If you take the Fourier transform, you assume that the nucleus doesn't move during scattering (the energy of the recoiled nucleus is negligible). This is correct at high Z (for example 208Pb) but it is incorrect for light nuclei, especially for proton.

I am not sure it's related to your question but the equivalence or not of the charge distribution and the form factor for defining the proton charge radius is addressed in this article: see G. Miller, Phys. Rev. C 99, 035202 (2019).

I think your understanding of the Breit frame is not correct. From https://en.wikipedia.org/wiki/Breit_frame, the Breit frame corresponds to the frame in which the momentum of a scattered particle is the opposite of its momentum before scattering, it doesn't impose $q^2 = -\mathbf{q}^2$. $q^2$ is equal to $ -\mathbf{q}^2$ if the mass of the nucleus is infinite and we actually have $q^2 = \omega^2-\mathbf{q}^2$ where $\omega$ is the energy of the recoiled nucleus.

1. A photon is an elementary particle so there is no form factor. Moreover, on-shell doesn't mean that $q^2=(p-p^\prime)^2=0$ but $p^2 = 0$.

2. & 3. If you take the Fourier transform, you assume that the nucleus doesn't move during scattering (the energy of the recoiled nucleus is negligible). This is correct at high Z (for example $^{208}$Pb) but it is incorrect for light nuclei, especially for proton.

I am not sure it's related to your question but the equivalence or not of the charge distribution and the form factor for defining the proton charge radius is addressed in this article: see G. Miller, Phys. Rev. C 99, 035202 (2019).

I am not an expert but I think your understanding of the Breit frame is not correct. From https://en.wikipedia.org/wiki/Breit_frame, the Breit frame corresponds to the frame in which the momentum of a scattered particle is the opposite of its momentum before scattering.

1. A photon is an elementary particle so there is no form factor.

2 & 3. If you take the Fourier transform, you assume that the nucleus doesn't move during scattering (the energy of the recoiled nucleus is negligible). This is correct at high Z (for example 208Pb) but it is incorrect for light nuclei, especially for proton.

I am not sure it's related to your question but the equivalence or not of the charge distribution and the form factor for defining the proton charge radius is addressed in this article: see G. Miller, Phys. Rev. C 99, 035202 (2019).

I think your understanding of the Breit frame is not correct. From https://en.wikipedia.org/wiki/Breit_frame, the Breit frame corresponds to the frame in which the momentum of a scattered particle is the opposite of its momentum before scattering, it doesn't impose $q^2 = -\mathbf{q}^2$. $q^2$ is equal to $ -\mathbf{q}^2$ if the mass of the nucleus is infinite and we actually have $q^2 = \omega^2-\mathbf{q}^2$ where $\omega$ is the energy of the recoiled nucleus.

1. A photon is an elementary particle so there is no form factor.

2. & 3. If you take the Fourier transform, you assume that the nucleus doesn't move during scattering (the energy of the recoiled nucleus is negligible). This is correct at high Z (for example 208Pb) but it is incorrect for light nuclei, especially for proton.

I am not sure it's related to your question but the equivalence or not of the charge distribution and the form factor for defining the proton charge radius is addressed in this article: see G. Miller, Phys. Rev. C 99, 035202 (2019).