When dealing with phonons and specific heat of solids, it seems the really important quantity to obtain is the density of states $N(\omega)$. When we have it, we can find the internal energy as
$$U(T)=\int N(\omega) \dfrac{\hbar \omega}{e^{-\hbar\omega/k_BT}-1}d\omega,$$
and having it we can also find the specific heat
$$C(T)=\dfrac{\partial U}{\partial T}.$$
Now, the way to find $N(\omega)$ is usually this: if the real lattice has primitive cell with volume $V$, the volume of the primitive cell of the $k$-space is $(2\pi)^3/V$. This means that, since there's just one point of the Bravais lattice per primitive cell, there are $V/(2\pi)^3$ points of the $k$-lattice per unit volume.
This in turns leads to integrals of the form
$$N(\omega)d\omega=\dfrac{V}{(2\pi)^3}\int d\mathbf{k},$$
where the integral is taken over the region between $\omega $ and $\omega+d\omega$, or equivalently
$$N(\omega)=\dfrac{V}{(2\pi)^3}\int\dfrac{dS}{|\nabla_\mathbf{k}\omega|},$$
where the integral is taken over the surface $\omega$.
This all is fine, given one dispersion relation $\omega(\mathbf{k})$ we can find $N(\omega)$ using those integrals, and with $N(\omega)$ we can find $U(T)$ and hence $C(T)$.
On the other hand, what if the basis of the Bravais lattice has more than one atom? For instance, a 2 atom basis?
This is quite common, but I'm failing to see how this affects all of this derivation. The naive guess would be that $N(\omega)$ would be multiplied by $2$, but this is just a guess. So, how the number of atoms in the basis affects this reasoning, and hence, the thermodynamic properties of a crystal, like specific heat?