thermodynamics - Collision Term in the Classical Boltzmann Transport Equation

I cannot get over the feeling that in the classical derivation of the collision term of Boltzmann's transport equation molecules that are already knocked out of a $(\textbf r, \textbf v)$ space volume are double-counted many times. In essence, my question really is: how can one multiply the number of "bullets" (incoming particles) by the number of "targets" (particles within an infinitesimal spatial-velocity element) to get the number of times a bullet hits a target? Isn't it that one "bullet" can only hit one "target"?

Please allow me to elaborate.

In the classical Boltzmann equation with the 2-body collision term for a gas of a single kind of molecule, the collision term is given as the sum of a gain through collision term and a loss through collision term.

Largely following Huang's Statistical Mechanics, the derivation of the loss term (the rate of decrease of particle density $f(\textbf r, \textbf v, t)$ owing to collisions) is reproduced as follows:

To a molecule in a given spatial volume $d^3 r$ about $\textbf r$, whose velocity lies in $d^3 v_1$ about $\textbf v_1$, other molecules of any given velocity $\textbf v_2$ in the same spatial volume pose as an incident beam. The flux of this incident beam is
$$I = |\textbf v_1 - \textbf v_2| f(\textbf r, \textbf v_2, t) d^3 v_2$$

So far so good. But the standard derivation goes on to say that the number of collisions happening in this spatial volume $d^3 r$ during $\delta t$ of the type $\{\textbf v_1, \textbf v_2\} \rightarrow \{\textbf v'_1, \textbf v'_2\}$ is given by
$$I \sigma(\textbf v_1, \textbf v_2 ; \textbf v'_1, \textbf v'_2) d^3 v'_1 d^3 v'_2 \delta t = |\textbf v_1 - \textbf v_2| f(\textbf r, \textbf v_2, t) \sigma(\textbf v_1, \textbf v_2 ; \textbf v'_1, \textbf v'_2) d^3 v_2 d^3 v'_1 d^3 v'_2 \delta t$$ A comment here from me before continuing on with the derviation: a single collision will knock the molecule with velocity $\textbf v_1$ out of the phase-space volume of $d^3 r d^3 v_1$ around $(\textbf r, \textbf v_1)$, how can it still be at the same location $\textbf r$ and velocity $\textbf v_1$ for the other collisions of the same type $\{\textbf v_1, \textbf v_2\} \rightarrow \{\textbf v'_1, \textbf v'_2\}$?

To help drive my point home, let's follow through with the standard derivation: The total rate of loss is obtained by integrating over all $\textbf v_2$, $\textbf v'_1$ and $\textbf v'_2$ and then multiplying the result by the number of molecules within the volume of $d^3 r d^3 v_1$ around $(\textbf r, \textbf v_1)$, i.e.,
$$df|_{\text{loss}} \equiv f(\textbf r, \textbf v_1, t+\delta t) d^3 r d^3 v_1 - f(\textbf r, \textbf v_1, t) d^3 r d^3 v_1 \\ = f(\textbf r, \textbf v_1, t) d^3 r d^3 v_1 \iiint |\textbf v_1 - \textbf v_2| f(\textbf r, \textbf v_2, t) \sigma(\textbf v_1, \textbf v_2 ; \textbf v'_1, \textbf v'_2) d^3 v_2 d^3 v'_1 d^3 v'_2 \delta t$$

To me, this further indicates that each molecule within $d^3 r d^3 v_1$ around $(\textbf r, \textbf v_1)$ undergoes $n$ number of collisions where $n = \iiint |\textbf v_1 - \textbf v_2| f(\textbf r, \textbf v_2, t) \sigma(\textbf v_1, \textbf v_2 ; \textbf v'_1, \textbf v'_2) d^3 v_2 d^3 v'_1 d^3 v'_2 \delta t$.

How are these multiple collisions possible? Isn't this double-counting?

The above is reproduced based on the derivation given in Huang's Statistical Mechanics, essentially same derivations can be found from multiple sources online, e.g., https://farside.ph.utexas.edu/teaching/plasma/Plasma/node33.html