Wikipedia states that Newton's second law of motion is:
At any instant of time, the net force on a body is equal to the body's acceleration multiplied by its mass or, equivalently, the rate at which the body's momentum is changing with time.
In other words, it is either
$$ F = ma $$
or
$$ F = \frac{d}{dt}(mv) = v\frac{dm}{dt} + m\frac{dv}{dt}. $$
These are definitely not equivalent. The first says that a changing mass does not affect the force. The second says it does.
So, which is it?
Your product rule is incorrect without being handled properly. The expression $F=v\frac{\text dm}{\text dt}+m\frac{\text dv}{\text dt}$ does not follow Galilean relativity; $v$ is not the same for inertial observers moving relative to each other, but we should have force being the same for all inertial observers.
You have to consider the rate of change in momentum of the entire system (the "original" system and the mass that causes the change in mass of the "original" system). If you handle this correctly for, say, a system of mass $m$ with velocity $\mathbf v$ colliding with a change in mass $\text dm$ with velocity $\mathbf u$, then we end up with
$$\mathbf F_\text{ext}+\mathbf v_\text{rel}\frac{\text dm}{\text dt}=m\frac{\text d\mathbf v}{\text dt}$$
where $\mathbf v_\text{rel}$ is the relative velocity $\mathbf u-\mathbf v$ between the system and added mass.
I think we should defend $F=ma$ more here though. If two masses collide and stick together, for example, you can either treat it as
Of course, this breaks down / becomes unreasonable when approximating the change in mass as continuous, such as in rocket propulsion, but still, saying $F=ma$ is incorrect because we can have variable-mass systems would be like saying Newtonian mechanics is incorrect because of special relativity. Sure, you are technically correct, but practically it's not too relevant.
Although mathematically inequivalent, the two definitions are equivalent within the scope of Newtonian Mechanics.
To understand the previous statement, one must understand the meaning of the quantities appearing in Newton's Second Law. In particular, definitions of momentum and mass are important.
Within Classical Mechanics, a point-like body (i.e. a body whose dynamical state is described only by its position and velocity) is characterized by a positive constant called mass. The operative definition of the mass can be based on the ratio of two bodies' accelerations or on the conservation of momentum. From the formal point of view, mass can be defined either from the analysis of the effect of the interaction between two isolated bodies (non-interacting with other bodies): $$ \frac{m_1}{m_2} = \frac{a_2}{a_1}, $$ where $a_i$ is the modulus of the acceleration of the $i$-th body, or from the requirement that the quantity $$ m_1 {\bf v}_1+m_2 {\bf v}_2, $$ for the same system, is constant. In both cases, there is an implicit assumption of constant mass. After the definition of mass, momentum is defined as the product of mass by velocity.
Such apparent limitation is not as severe as it could seem because, starting from the formulation of the equations of motion for constant-mass bodies, it is possible to extend the theory to treat systems of bodies if some of these compound systems vary their total mass. This is the only safe way to extend Newtonian mechanics to the cases of so-called mass-varying systems. A blind use of $$ \frac{{\mathrm d}{\bf p}}{{\mathrm d}t} = \frac{{\mathrm d}{ (m{\bf v}})}{{\mathrm d}t} = m \frac{{\mathrm d} {\bf v}}{{\mathrm d}t} + \frac{{\mathrm d}{m}}{{\mathrm d}t} {\bf v} $$ is well known to be wrong in many cases of mass-varying systems.
They are equivalent, because it is assumed (in both) that the body does not lose or gain any massive parts.
In case the body is losing massive parts, such as a rocket, both formulations seem inadequate, because it is not clear what the system considered when using words "mass" or "momentum" and "net force" is - whether the force of the expelled gases on the rocket motor nozzles is to be accounted in the net force, and whether $m$ is mass of and inside the rocket, or mass of the whole system, including the expelled gases.
But in fact, the $\mathbf F=m\mathbf a$ formulation is correct and applicable to the rocket, provided $m$ is the mass of and inside the rocket, not including the expelled gases (thus $m$ is changing in time), and $\mathbf F$ includes the force on the nozzles due to expelled gases. This force can be expressed as $\frac{dm}{dt} \mathbf c_0$ where $\mathbf c_0$ is the final velocity of the expelled gases in the frame of the rocket. So the equation of motion for the rocket which experiences external force $\mathbf F_{ext}$ (e.g. due to atmosphere) is
$$ m\mathbf a = \mathbf F_{ext} + \frac{dm}{dt} \mathbf c_0 $$
and this equation holds in all reference frames.
The other formulation $\mathbf F = d\mathbf p/dt$, if $\mathbf p$ means momentum of the rocket $\mathbf p=m\mathbf v$, is incorrect in general, because due to the term $\frac{dm}{dt}\mathbf v$, it depends on the frame of reference. This is not acceptable, because force does not depend on frame of reference - the "$dp/dt$ formulation" would lead to
$$ m\mathbf a + \frac{dm}{dt}\mathbf v = \mathbf F_{ext} + \frac{dm}{dt} \mathbf c_0. $$
This equation is correct only in the inertial frame where the rocket has zero velocity, where the term $\frac{dm}{dt}\mathbf v$ is zero.
For a collection of particles, you have the definition of momentum $$ \boldsymbol{p} = \sum_i m_i \boldsymbol{v}_i \tag{1}$$
for which you can show that for a rigid body there exists a point C riding along the body, the center of mass, such that $$ \boldsymbol{p} = m \boldsymbol{v}_C \tag{2}$$ with $m = \sum_i m_i$ and the assumption that mass is constant.
Now Netwton's 2nd law relates the net force acting on a body with the change in momentum
$$ \boldsymbol{F}_{\rm net} = \tfrac{\rm d}{{\rm d}t} \boldsymbol{p} \tag{3} $$
The above is always the definition and if you do not assume a rigid body then (1) leads to
$$ \boldsymbol{F}_{\rm net} = \tfrac{\rm d}{{\rm d}t} \left( \sum_i m_i \boldsymbol{v}_i \right) = \sum_i \tfrac{\rm d}{{\rm d}t} \left( m_i \boldsymbol{v}_i \right) = \sum_i \left( (\tfrac{\rm d}{{\rm d}t} m_i) \boldsymbol{v}_i + m_i \boldsymbol{a}_i \right) \tag{4} $$
But if you only have one particle, or all the particles belong to the same rigid body you get
$$ \boldsymbol{F}_{\rm net} = \tfrac{\rm d}{{\rm d}t} \left( m \boldsymbol{v}_C \right) = m \tfrac{\rm d}{{\rm d}t} \left( \boldsymbol{v}_C \right) = m \boldsymbol{a}_C \tag{5} $$
You can say that both (4) and (5) are correct within their own assumptions, as they both derive from the 2nd law of motion (3).
