Rather than call it a variable mass system consider the problem as a flow of mass with the system under consideration having a constant mass.
Suppose there is a system which has a momentum $\vec p_{\rm system}(t)$ at a time $t$ and a momentum $\vec p_{\rm system}(t+\Delta t)$ at a time $t+\delta t$.
Applying Newton's second law with an external force $\vec F_{\rm external}$ acting on the system gives
$$\vec F_{\rm external}=\frac {d\vec p_{\rm system}}{dt} = \lim \limits_{\Delta t \to 0}\frac {\Delta\vec p_{\rm system}}{\Delta t} = \lim \limits_{\Delta t \to 0}\frac {\vec p_{\rm system}(t+\Delta t) = \vec p_{\rm system}(t)}{\Delta t}$$
Note that the system has constant mass.
So how to deal with, say, the rocket?
As was stated in the answer to the previous question $$\vec F_{\rm external} = \frac{d\vec p}{dt} = \frac{d}{dt}(m\vec v) = \frac{dm}{dt}\vec v + m\frac{d\vec v}{dt}$$
might be a perfectly valid mathematical differentiation but is not good Physics.
For example what is $\vec v$ in the term $\dfrac{dm}{dt}\vec v$?
It is the velocity of the rocket.
A reason why the differentiation of $mv$ with respect to time seems to be an easy way to obtain the rocket equation is that when Newton's second law is used correctly with a system of constant mass the equation is
$$\vec F_{\rm external} = \frac{dm_{\rm rocket}}{dt}\vec u + m_{\rm rocket}(t)\frac{d\vec v_{\rm rocket}}{dt}$$ where $\vec u$ is the velocity of the exhaust gases relative to the rocket and $\dfrac{dm_{\rm rocket}}{dt}$ is negative.
This equation looks very similar to the previous one except that the velocity of the rocket $\vec v$ in that equation has been replaced by another velocity in the correct equation - the velocity of the exhaust gases relative to the rocket $\vec u$.
The correct equation is then often rewritten as $$\vec F_{\rm external} - \frac{dm_{\rm rocket}}{dt}\vec u = m_{\rm rocket}(t)\frac{d\vec v_{\rm rocket}}{dt}$$
where the term $- \dfrac{dm_{\rm rocket}}{dt}\vec u $ is called the thrust and has a positive contribution to the left hand side as $\dfrac{dm_{\rm rocket}}{dt}$ is a negative quantity.
If there are no external forces acting on the system, $\vec F_{\rm external}=0$, then the centre of mass of the system (rocket & exhaust gases) does not change position.
The application of an external force means the the centre of mass of the system accelerates but what happens to the exhaust gases is of no interest as it the velocity of the rocket which you are trying to find.
If there are no external forces acting and $\hat v_{\rm rocket}= - \hat u$ then one has to solve the differential equation $$- \frac{dm_{\rm rocket}}{dt}\, u = m_{\rm rocket}(t)\frac{d v_{\rm rocket}}{dt}$$ to get the result $v_{\rm rocket,final} =v_{\rm rocket,initial} +u\, \ln\left(\dfrac {m_{\rm rocket,initial}}{m_{\rm rocket,final}}\right) $
and with a gravitational force acting in the opposite direction to the velocity of the rocket then the differential equation to be solved is
$$-m_{\rm rocket}(t)g- \frac{dm_{\rm rocket}}{dt}\, u = m_{\rm rocket}(t)\frac{d v_{\rm rocket}}{dt}$$
and the solution for the velocity has an extra term $-g(t_{\rm final}-t_{\rm initial})$ in it as a result of the action of the gravitational force.
