Boltzmann statistics, or rather, the Boltzmann distribution, describes a system in thermodynamical equilibrium.
For a system that may be in non-equilibrium its one-particle distribution function $f(\mathbf x, \mathbf v,t)$ obeys the Boltzmann equation $$\frac{\partial f}{\partial t} + \mathbf v \cdot \nabla_x f + \frac{\mathbf F}{m} \nabla_p f = g$$
where $f(\mathbf x, \mathbf v, t)d\mathbf x d\mathbf v$ is the probability density for particles at $\mathbf x$ with velocity $\mathbf v$ at time $t$, and $\mathbf F$ is the force and $g$ represents collisions. One can check that the Boltzmann distribution $f = \exp(-mv^2/2 - U)$ where $U$ is the potential $\mathbf F = -\nabla U$ is a static solution for $g = 0$.
The term $g$ comes from correlations between particles, or that the probability to find one particle at $1$ and another at $2$ is not just the product of $P(1)$ and $P(2)$. Because of this correlation the one-particle distribution function does not evolve simply according to the forces produced self-consistently by the particles.
The $\mathbf F$ term represents the total force on one particle from all other particles, whereas the correlation term $g$ represents two-particle interactions. (This statement is made more precise below.) Thus $g$ is called the collision term - it stands for two particles "colliding" as opposed to particle motion in self-consistent fields produced by all particles.
The term collision is indeed a little misleading as these do not need to be hard spheres colliding. That the collision term appears is independent of the details of the particle interactions, but of course its details will vary. In particular its magnitude may vary with the strength of interactions. A gas in a box is certainly much more dense and interacts more strongly than a galaxy (gravitation is very weak), motivating that the collision term is not important for galaxies.
In more detail, the function $f$ is a one-particle distribution function, but the system consists of $N \gg 1$ particles. To specify an ensemble we really need an $N$-particle distribution function, $f_N(\mathbf x_1, \mathbf v_1, \ldots, \mathbf x_N, \mathbf v_N, t)$. For reasonable physics (more precisely: Hamiltonian systems) phase space density is conserved along the motion. In an equation, $$0 = \frac{\partial f_N}{\partial t} + \sum \dot{\mathbf x}_i \cdot \nabla_{x_i} + \sum \dot{\mathbf a}_i \cdot \nabla_{v_i} f_N$$
by applying the chain rule. Of course, $\dot x = v$ and $\dot v = a = F/m$.
From the $N$-particle distribution function we can find an $(N-1)$-particle distribution function by integrating over $\mathbf x_N$ and $\mathbf v_N$. It is then possible to find an equation for $\partial f_{N-1}/\partial t$. Naturally this equation will have a term with an integral involving $f_N$. This procedure can be repeated $N-1$ times to find an equation for the one-particle distribution $f_1$. This equation is $$0 = \frac{\partial f_1}{\partial t} + \mathbf v\cdot \nabla_x f_1 + \frac{N -1}{V} \int \mathbf a_{12} \cdot \nabla_{v_1} f_2(\mathbf x_1, \mathbf v_1, \mathbf x_2, \mathbf v_2)\, d\mathbf x_2 d\mathbf v_2$$
where $V$ is the volume of the system and $\mathbf a_{12}$ is the acceleration of a particle at $(\mathbf x_1, \mathbf v_2)$ due to a particle at $(\mathbf x_2, \mathbf v_2)$. This hierarchy of equations for $f_N$ through $f_1$ is called the BBGKY hierarchy.
To get to the Boltzmann equation we let $f_2(1,2) = f_1(1)f_1(1) + g(1,2)$ where the function arguments have the obvious interpretation. The function $g$ measures how the two-particle distribution is not just the product of two one-particle distributions, or how particles are correlated. Inserting this cluster expansion in the BBGKY equation for $f_1$, we get $$0 = \frac{\partial f}{\partial t} + \mathbf v\cdot \nabla_x f_1 + (\nabla_{v} f_1 ) \cdot \int \mathbf a_{12} f_1 (\mathbf x_2, \mathbf v_2, t) d\mathbf x_2 d\mathbf v_2 + \tilde g$$
where $\tilde g$ is a term involving $g$. This is on the same form as the Boltzmann equation above, taking $F$ to be the force produced by the particles self-consistently.
