Expanding Newton's Second Law and changing $v$ to $V_e$ gives you the rocket equation. Where does this change of $v\to V_e$ come from? I've noticed that from Newton's Second Law, if you expand the right side using the product rule and change the v to Ve, the rocket equation appears very easily...  Why is this second term of velocity considered to be the Exhaust Velocity when Newton's law only started off with m*v?

What I'm trying to figure out is how you know that second term is referring to the velocity of the expelled mass.
F = dp/dt = d(mv)/dt = m * dv/dt + v * dm/dt
As far as I can tell we started with p of the rocket, which is m*v, and that's expanded to m * dv/dt + v * dm/dt just by using the product rule, so how come that second term's v is referring to the expelled mass even though it just came from the expansion of the rocket's momentum?
 A: $$\vec F=\frac{d \vec p}{dt}=m\frac{d \vec v}{dt}+\vec v\frac{dm}{dt}.$$
Presumably the equation is supposed to give the resultant force on a body of varying mass, the rocket, though I don't see, in that case, why you would equate the force to zero. If the equation is to be applied to the complete system of rocket and exhaust gases, then $\vec F$ does equal zero (in a field-free region), but so too does $\frac{dm}{dt}$.
In Newtonian physics, corporeal mass is conserved: a body can't lose mass without that mass constituting another 'body'. The force on the rocket depends on the change in velocity of that body as it leaves the parent body (the rocket). The equation I've quoted at the top can't be right for the rocket because it doesn't take this velocity change into account. I'd argue that we can't apply Newton's second law to a body of varying mass as such. This may be controversial, but what is certainly true is that we can apply Newtonian principles to two bodies of fixed mass: the exhaust gas that leaves in a certain time, and the rest of the rocket (including exhaust that is yet to leave). Here's what I mean...
Let $\mu$ be the mass of exhaust gases leaving the rocket per unit time and $\vec {v_E}$ be the ejection velocity of the gases relative to the rocket. But in any time interval $\Delta t$,
momentum gained by gases + momentum gained by rest of rocket ($m$) = 0
So
$$\mu \Delta t\ \vec {v_E} + m\Delta \vec v =0$$
So in the limit as $\Delta t$ approaches zero
$$-\mu \vec{v_E} =m\frac{d\vec v}{dt}\ \ \ \ \ \ \ \text{that is}\ \ \ \ \ \ \ \frac{dm}{dt}\vec{v_E} =m\frac{d\vec v}{dt} $$
Or in terms of speeds, that is the magnitudes of the velocities (the velocities of rocket and exhaust gas being in opposite directions),
$$\frac{dm}{dt} v_E =- m\frac{dv}{dt}$$
We can now see what's going on. $-\frac{dm}{dt}\vec {v_E}$ is the rate of change of momentum of the exhaust gases, so equal and opposite to the force on the rocket. There is no term in $\frac{dm}{dt}\vec v$, (though it would appear in our final equation if we expressed the velocity of exhaust gases not relative to the rocket, but as a (varying) absolute velocity in the frame of reference in which the rocket's velocity is $\vec v$).
The rocket equation that we've finished up with can indeed be generated by replacing $v$ with $v_E$ and putting $F=0$ in the 'product derivative' formula that you started with, but this is little more than a co-incidence. It is not a sound derivation.
