Why do neutron stars with more mass have smaller volume?
The key point is that the Heisenberg principle restricts the number of particles in a (hyper)-volume of phase space rather than a volume of plain 'ole physical space.
That is, there can be at most two neutrons in a (hyper-)box defined by a volume in physical space times a volume in momentum space.
But the kinetic energy of a particle depends on it's momentum.
Neutron stars stop collapsing when, in order to fit all the particles into less physical space you would have to give some of them enough more momentum that the change costs more added kinetic energy that you get back in gravitational energy.
When you add more mass the amount of gravitational energy you get from a particular change in radius increases, so you can afford to move neutrons to higher momentum in order to fit them into less physical space.
You can actually see a similar phenomena at work in the sizes of some atoms: higher Z atoms (with more electrons) can take up less space than lower Z atoms (with fewer). Because more protons in the nucleus means you get more energy from moving the electrons inward and can therefore afford to have them at higher momentum.
Try this argument.
To be in hydrostatic equilibrium, the pressure gradient inside a star must equal (minus) the density multiplied by the gravitational field $$ \frac{dP}{dr} = - \rho g$$
If we take the average pressure gradient to be $-P_c/R$, where $R$ is the stellar radius and $P_c$ is the central pressure, then this will roughly be equal to average density multiplied by the average gravitational field. So in proportionality terms $$ \frac{P_c}{R} \sim \frac{M}{R^3}\frac{M}{R^2}\, ,$$ where $M$ is the stellar mass and thus $$ P_c \sim M^2 R^{-4} \tag*{(1)}$$
The relationship between mass and radius will depend on what provides the pressure.
If it is perfect gas pressure (in a non-degenerate star), then $P_c \propto \rho T \sim MT/R^3$. But in a main sequence star, the core temperature is roughly fixed, because hydrogen burning has a strong temperature dependence and hardly changes as the mass changes. Thus we have from eqn (1): $$ P_c \sim MR^{-3} \propto M^2 R^{-4}$$ $$ \rightarrow \ \ \ \ R \propto M $$
Now consider (non-relativistic) degeneracy pressure. This scales as $\rho^{5/3}$ and is independent of temperature. Thus $P_c \propto M^{5/3} R^{-5}$. Putting this into eqn (1): $$P_c \sim M^{5/3} R^{-5} \propto M^2 R^{-4}$$ $$ \rightarrow \ \ \ \ R \propto M^{-1/3} $$
Note that real neutron stars are not governed by the ideal equation of state for degenerate neutrons. Neutrons in fact are strongly intercting particles when compressed to separations of $\sim 10^{-15}$ m. The interaction is repulsive and leads to a "hardening" of the equation of state, such that $P_c \sim \rho^2 \propto M^2 R^{-6}$. If we put this into equation (1) we find that there is only one value of $R$ that will satisfy the equation. i.e. That the radius does not depend on mass. If you look at many examples of theoretical mass-radius relations for neutron stars you will see that there usually is a range of masses for which the radius is nearly constant.