Relation between freezing point depression and vapor pressure
A freezing point is a temperature at which solid and liquid phases are in equilibrium with each other. In equilibrium the chemical potentials of each component in all phases present are equal. It means that if there are three phases in the system, solid, liquid and gas, then the chemical potential should be equal in all the three. As a consequence, the vapor pressure of a given component over the liquid phase should equal that over the solid phase. But if there are only solid and liquid phases, I see no need to involve gas. The basic derivation, which I was taught, of the freezing point depression of liquid solvent A (when the solute is insoluble in solid A) starts with $$\mu_\mathrm{A}(\mathrm{solid})=\mu_\mathrm{A}(\mathrm{solution})=\mu_\mathrm{A}(\mathrm{pure\,liquid})+RT\ln{x_\mathrm{A}}$$ - no gases here.

I might miss something here, but my guess is that your lab manual is just trying to be more obscure than it ought to be (no offence to its authors intended). While the expression for $\mu_\mathrm{A}(\mathrm{solution})$ itself can be derived from the Raoult's law, which requires gases, that's slightly different, although related, story. Provided that the expression for $\mu_\mathrm{A}(\mathrm{solution})$ is already known, I can see no sound reason for inviting the vapor pressures to explain cryoscopy.