This is from pages 32-33 of Physics for Mathematicians. A rocket + fuel system of mass $m(t)$ is moving along at velocity $v(t)$, with its fuel being ejected at the velocity (relative to the rocket) $q(t)$.
In a short time interval $[t, t + h]$, the amount of fuel ejected is $m(t) - m(t + h)$, and therefore the momentum of expelled fuel will be close to $[m(t) - m(t+h)]q(t)$. Thus, the momentum of the fuel in the other direction, pushing the rocket forward, will be the negative of this.
How do we deduce that the momentum of the fuel remaining in the rocket is the negative of the momentum of the fuel flying out of the rocket? This could be justified if we assumed that there are no external forces on the fuel, and that the fuel is initially at rest in the rocket, but I'm not sure about either of those assumptions.
...so the force on the rocket must be the derivative, $m'(t)q(t)$...
Why? All I can recognize is that $m'(t)q(t)$ is, given the expressions mentioned previously, the derivative of the momentum of the expelled fuel at time $t$. This is the force on the expelled fuel, which by what was said before is the negative of the force on the fuel pushing into the rocket, so we're saying
Force on fuel pushing into rocket = - Force on rocket
Which seems kind of like Newton's third law, but who says that the force on the left hand side is being applied by the rocket?