Saying that warm air "holds" more moisture is technically incorrect, but is a common colloquialism. Let's break it down to the technicalities.
Let's consider a glass of water with a vacuum (no air) above it. What will happen? The molecules that are at the top most layer of the water will evaporate. At what rate will the water evaporate? Better yet, what is evaporation?
Evaporation is when the water molecules gain enough kinetic energy (how fast they vibrate) to break the bonds that hold them to one another. Kinetic energy is dependent on temperature. So the molecules vibrate faster, break their bonds, and enter the vacuum as a vapor. Some molecules will stay as a vapor in the vacuum, but others will reenter the liquid. When the molecules enter the liquid as fast as they are leaving, then it is saturated.
If the air is cooled down, then the rate at which molecules leave the liquid slows down. The molecules entering the liquid do not slow down at the same rate, causing the liquid to grow toward it's initial state.
Note that I specifically said it is a vacuum. Instead of a glass of water, picture the water as little drops. The atmosphere can act to warm or cool these drops, and vice-versa.
In the more nitty-gritty aspect of this, the equation that describes the vapor pressure as a function of temperature is called the Clausius-Clapeyeron equation/relation. The American Meteorological Society has one approximate solution, but I prefer this equation: $$e_{sat}(T)=611 Pa \exp[\frac{L_v}{R_v}(273.15^{-1}-T^{-1})]$$, where $L_v$ is the latent heat of vaporization, $R_v$ is the specific gas constant for water vapor, and $T$ is the absolute temperature in Kelvin. Combined with the ideal gas law for water vapor (assuming saturation) $$e_{sat}(T)V=m_vR_vT$$, and given the volume ($V$) we can write an expression for the mass of water vapor $m_v$. The equation comes out to $$m_v=611 Pa \exp[\frac{L_v}{R_v}(273.15^{-1}-T^{-1})]V R_v^{-1}T^{-1}$$
To answer your final question, the molecules are approximated as being infintessimally small, per the ideal gas law. To be more specific, one molecule of water is about 7.08$\times$ 10$^{-19}$ cubic feet (after some math), so the added volume is considered negligible. In short, the molecules are treated as point masses.