There are many questions here so I will answer some more fully than others.
This Hamiltonian comes from the atom-field interactions and atom-atom interactions. Consider that there are a bunch of atoms, each labeled by $i$, at different locations. If light resonant with the energy difference between the ground state $|g_i\rangle$ and the [highly excited] Rydberg state $|r_i\rangle$ is shone on the $i$th atom, the optical Bloch equations indeed say that the interaction Hamiltonian will be of the form $\Omega_i(|r_i\rangle\langle g_i|+|g_i\rangle\langle r_i|)$. This comes from the electronic dipole moment interacting with a quantized field mode via the Jaynes-Cummings Hamiltonian, taken to the limit of a classical driving field. Since there can be a different laser with a different intensity driving each particular transition for each particular atom, the interaction strengths $\Omega_i$ may differ, so they each get a unique label $i$. Note that $\sigma_x^i=|r_i\rangle\langle g_i|+|g_i\rangle\langle r_i|$, so we have accounted for the first sum.
Next, consider that the driving fields are no longer on resonance with the energy difference between the Rydberg and ground states. Say the driving field has frequency $\omega_{\mathrm{f}}$. Then, just like in the optical Bloch equations or the Jaynes-Cummings model, the interaction picture (the frame rotating at the frequency of the driving field) retains a term from the atoms' free evolution (which would have looked something like $\omega_{i}(|r_i\rangle\langle r_i|-|g_i\rangle\langle g_i|)/2$) when $\Delta_i=\omega_{\mathrm{f}}-\omega_{i}\neq 0$. We can always change where the zero of energy is such that the atom's free evolution looks like $\omega_{i}|r_i\rangle\langle r_i|=\omega_i n_i$; then, in the interaction picture known as the rotating frame, the remaining energy imbalance continues to add an extra phase to the excited state via the term $(\omega_i-\omega_{\mathrm{f}})n_i=-\Delta_i n_i$. Since the detunings may be different for each atom, this accounts for the second term. The detuned version of the interaction also changes the Rabi frequencies $\Omega_i$ in a known way, but the physics is the same; this is all the same as the optical Bloch equations but with lots of atoms each interacting with a light field.
Reminder that this is a Hamiltonian, not a measurement operator. The Hamiltonian gets applied to the state in an exponential. It is true that $\exp(i \theta \sigma_x)$ will cause oscillations between two states, but it is not true that $\exp(i \theta n)$ will cause a projection. Instead, we have that $\exp(i \theta n_i)(\alpha |g_i\rangle+\beta|r_i\rangle)=\alpha |g_i\rangle+\beta e^{i\theta}|r_i\rangle$. The "detuning" terms cause a relative phase to accumulate between the states $|r_i\rangle$ and $|g_i\rangle$. As for being a Hamiltonian, it simply means that states $|r_i\rangle$ have more energy than states $|g_i\rangle$.
The final term is the most familiar, yet is the most mysterious. This is the Coulomb interaction between two electrons! When you do a multipole expansion of the interaction between two atomic dipoles, the most important term is a dipole-dipole interaction, where any two electrons that are both in the Rydberg state will repel each other if they are close enough together. You can read more details in, e.g., this paper. When neighbouring electrons are not fully excited, they do not repel each other as much. So, if the interaction energy between a pair of excited electrons is $V_{ij}$, then this term will only contribute if both atom $i$ and atom $j$ have their electrons in the Rydberg state. That is what the operator $n_i n_j$ does: it only acts in a non-trivial way on states where both atoms $i$ and $j$ are in their Rydberg states. We have that $$(n_i n_j) (|r_i\rangle |r_j\rangle)\neq 0,$$ but $$(n_i n_j) (|g_i\rangle |r_j\rangle)=(n_i n_j) (|r_i\rangle |g_j\rangle)=(n_i n_j) (|g_i\rangle |g_j\rangle)= 0.$$ The actual format of the $V_{ij}$ operator depends on the distance between the $i$th and the $j$th atom and is often due to the van der Waals interaction.
Finally, what about ratios. We know that detuning can be positive or negative depending on whether the frequency of the field is less than or greater than the atomic energy difference. In many cases, the atoms all have similar detunings and the field strenght is similar across all of the atoms, so we can drop the index $i$ on the Rabi frequency and the detuning to just consider the parameters $\Omega$ and $\Delta$. It is of course reasonable to consider both parameters, but people like to use the ratio, to have a shorthand for the situation in which $\Omega \gg |\Delta|$ (then $|\Omega/\Delta| \gg 1$) or $\Omega \ll |\Delta|$ (then $|\Omega/\Delta| \ll 1$), etc. There is no extra significance to the ratio, it is just often convenient to only consider one parameter instead of two! Note that this is a Hamiltonian, governing time evolution, so you can always multiply or divide by a constant and get the same time evolution, it will just happen slower or faster. So you can divide the Hamiltonian by $\Delta$ and get the exact same dynamics, just at a different rate. Then, the Hamiltonian will depend on $\Omega/\Delta$ explicitly, and if you know the dynamics will have most of the atoms go to their ground states then you have learned something "only" from the ratio.
The idea of the quoted sentence is that if there is a large negative detuning then the most significant contribution to the energy is the detuning term. Then, the ground state of the overall many-body interacting system is likely to be the same as the ground state of the detuning part of the Hamiltonian, which is the state where each atom is in its ground state. If there is a large positive detuning, then the Hamiltonian will have many terms like $-|\Delta||r_i\rangle\langle r_i|$, so the overall ground state (with lowest energy) will have a tendency to have more atoms in their Rydberg states than when the detuning was negative. One then has to consider that neighbouring Rydberg states repel each other, so the overall dynamics become interesting.
In sum: dipole-dipole interactions and dipole-field interactions for a quantized large dipole.
