Multi-particle state in relativistic quantum mechanics

I was following lecture notes of Sidney coleman (https://arxiv.org/abs/1110.5013)(page 7) where he stated:

"The addition of relativity is necessary at energies $E geq m c^{2}$. At these energies
$$
p+p rightarrow p+p+pi^{0}
$$
is possible. At slightly higher energies
$$
p+p rightarrow p+p+p+bar{p}
$$
can occur. The exact solution of a high energy scattering problem necessarily involves many particle processes.

You might think that for a given $E,$ only a finite number, even a small number, of processes actually contribute, but you already know from NRQM that that isn’t true.
$$
H rightarrow H+delta V quad delta E_{0}=langle 0|delta V| 0rangle+sum_{n} frac{|langle 0|delta V| nrangle|^{2}}{E_{0}-E_{n}}+cdotstag{1}
$$
Intermediate states of all energies contribute, suppressed by energy denominators. For calculations of high accuracy effects at low energy, relativistic effects of order $(v / c)^{2}$ can be included. Intermediate states with extra particles will contribute corrections of order
$$frac{E}{m c^{2}} = frac{text{Typical energies in problem}}{text{Typical energy denominator}} sim frac{m v^{2}}{m c^{2}}=left(frac{v}{c}right)^{2}tag{2}$$

As a general conclusion: the corrections of relativistic kinematics and the corrections from multiparticle intermediate states are comparable; the addition of relativity forces you to consider many-body problems. "

I have the following doubts regarding this paragraph:

1. How does the equation 1 and 2 related? Precisely why do we consider only the kinetic energy for the numerator(for interaction part in the perturbation expression) while the rest mass energy $mc^2$ for the energy denominator?

My thinking goes as follows: For any pair production En-E0>mc^2 hence the factor in the denominator. Am I right?