Answer
$H_0$ is the "non magnetic" Hamiltonian operator, which means the part which is not coupled to the magnetic field. In the purpose of the exercise, you can think the quantum number $\alpha$ as the set of momentum-spin $k,\sigma$. For instance if you have an electron system in a cubic lattice with hopping between nearest neighbours sites, you have $\varepsilon_{k\sigma} = -2t\sum_{\mu} \cos{(\vec{k} \cdot \hat{e}_{\mu})}$ (where $\vec{k}$ is the momentum and $\hat{e}_{\mu}$, $\mu=x,y,z$ is the versor of an axis). Anyway the explicit form is not necessary to solve the exercise.
Solution to your doubt
Your doubt is justified: if you want to give a full quantum treatment of the electron gas coupled to a magnetic field, you not only have to take into account the coupling with the spin $-\mu_B\vec{S}\cdot \vec{B}$, but also the coupling to the orbital angular momentum $-\mu_B\vec{L}\cdot \vec{B}$, which indeed affects $H_0$.
However one can prove that the magnetic susceptibility can be written as $\chi = \chi_L + \chi_S$, where $\chi_S$ is what you get if you consider the spin coupling only, while $\chi_L$ is what you get if you consider the orbital coupling only. A reference is Landau Vol. 5 (which is really precise, but a little tough to understand).
In conclusion, the spirit of the exercise is to compute only one of these two contributions, in particular $\chi_S$.
Curiosity
It turns out that $\chi_S$ is positive (paramagnetic term), while $\chi_L$ is negative (diamagnetic term), and that $\chi_L = -\chi_S/3$ if you assume no underlying lattice structure. In conclusion the free electron gas is paramagnetic, and the effect of the term you are neglecting is to reduce the paramagnetic response by a factor $2/3$.
