First, we should remove some misconceptions that are actually getting in the way of the discussion here. Newton's laws were not formulated for point masses! The very first two definitions in the Principia frame bodies as continua, irrespective of any distinction between composite versus elementary, and do so with the implied understanding that the same laws apply to a whole body as apply to each of its parts. It is for this reason that you also have the third law - as that is the key enabling condition that allows you to scale the first and second law up from a body's parts to the body as a whole.
Definition 1 states that the quantity of matter comprising a body (its "mass") is - what we would say in contemporary language - the integral of its density over its volume. In so stating, it asserts that mass is additive and (tacitly) that it is positive.
By "additivity" I will also be referring to the stronger sense where two bodies can occupy overlapping regions, such as with the mixing of fluids or the lodging of a soda straw into a tree trunk by a tornado.
Definition 2 associates each part of the body with a velocity, and states - as we would today say - that the quantity of motion of the body (its "momentum") is the integral of the product of the density and velocity over the volume of the body. This means that the momentum is also additive, and it also gives you a definition for the average velocity of a body and - by these means - the average position of a body, up to a constant of time integration. It tacitly states (by the context of subsequent discussion) that this constant (that is: the constant of integration of the time integral of the "quantity of motion") is also additive; specifically that mass-moment-at-a-fixed-time is additive.
This part-whole and upward-scalability approach, where no consideration is paid to "elementariness", in fact, might be seen as a precursor to the approach Wilson later adopted in Quantum Field Theory.
There is nothing said in Newton's formulation of his laws about point masses in any of this. The only assumption (tacitly) made, in carrying on to formulate the laws, is that a body has consistent contents - that its mass be constant. The assumption incurs no real loss of generality, since a body of variable mass can be treated as a component of a larger body of constant mass, where the choice of where to draw the dividing line between what comprises that body, versus what comprises the rest of the larger body it is contained in, is allowed to vary with time.
It might do well to review the reply I gave here Rocket Equation where I spell out the derivation in general terms. That's exactly the approach adopted there.
There's nothing wrong with continuing to assert $ = d/dt$, for a body with mass $m$ and momentum $$, as long as you understand that when the mass is variable, then $$ is no longer frame-independent. Under Galilean a boost by $$, the coordinates, coordinate differentials, velocity, partial derivative operators and moving time derivative transform as
$$(,t) → ( - t, t), \quad
(d,dt) → (d - dt, dt), \quad → - ,\\
\left(∇,\frac{∂}{∂t}\right) → \left(∇, \frac{∂}{∂t} + ·∇\right), \quad
\frac{d}{dt} → \frac{d}{dt},$$
where
$$ = (x, y, z), \quad ∇ = \left(\frac{∂}{∂x}, \frac{∂}{∂y}, \frac{∂}{∂z}\right), \quad \frac{d}{dt} = \frac{∂}{∂t} + ·∇.
$$
This is what leads to the transport laws for continuua:
$$m_Ω = \int_Ω ρ dV\quad⇒\quad\frac{dm_Ω}{dt} = \frac{d}{dt} \int_Ω ρ dV = \int_Ω \left(\frac{∂ρ}{∂t} + ∇·(ρ)\right) dV,$$
where the region $Ω$ comprising the body can be time-varying, where $ρ$ is the density within the region, $$ the associated velocity and $dV = dx∧dy∧dz$ is the volume 3-form. If the region follows the body's contents, then the mass is constant and that leads to the transport equation:
$$\frac{∂ρ}{∂t} + ∇·(ρ) = 0.$$
Otherwise if the contents are not constant, then you're treating this as part of a larger body where the boundary of $Ω$ is allowed to have flow in and out. And that leads to the transport law for momentum:
$$\frac{∂(ρ)}{∂t} + ∇·(ρ) = ,$$
where tensor-dyad notation is used, and the decomposition $ = ρ$ for (the additive) force density $$ into the product of mass density $ρ$ and force per unit mass $$ is frequently employed. As stated, it is not additive. Therefore, solely on consistency grounds, there needs to be an additional term $$ alongside the dyad $ρ$ that absorbs the non-additivity, thus leading to the modified equation
$$\frac{∂(ρ)}{∂t} + ∇·(ρ + ) = .$$
The dyad $$ is then identified as the body's stress tensor. Thus, for two bodies with respective densities, velocities and stress tensors $\left(ρ_0, _0, _0\right)$ and $\left(ρ_1, _1, _1\right)$ occupying respective regions $Ω_0$ and $Ω_1$ - that may overlap - you can treat them as being part of one body within a region $Ω ⊇ Ω_0 ∪ Ω_1$, by just requiring $\text{supp }ρ_0 ⊆ Ω_0$ and $\text{supp }ρ_1 ⊆ Ω_1$ and writing
$$
\int_{Ω_0} ρ_0 dV + \int_{Ω_1} ρ_1 dV = \int_Ω ρ dV,\\
\int_{Ω_0} ρ_0 _0 dV + \int_{Ω_1} ρ_1 _1 dV = \int_Ω ρ dV,\\
\frac{∂ρ_0}{∂t} + ∇·(ρ_0_0) + \frac{∂ρ_1}{∂t} + ∇·(ρ_1_1) = \frac{∂ρ}{∂t} + ∇·(ρ),\\
\frac{∂\left(ρ_0_0\right)}{∂t} + ∇·\left(ρ_0_0 + _0\right) +
\frac{∂\left(ρ_1_1\right)}{∂t} + ∇·\left(ρ_1_1 + _1\right) =
\frac{∂\left(ρ\right)}{∂t} + ∇·\left(ρ + \right),\\
_0 + _1 = ,
$$
then the additivity entails the following part-whole decompositions:
$$
ρ_0 + ρ_1 = ρ ⇒ ρ = ρ_0 + ρ_1,\\
ρ_0_0 + ρ_1_1 = ρ ⇒ = \frac{ρ_0_0 + ρ_1_1}{ρ_0 + ρ_1},\\
ρ_0_0 + _0 + ρ_1_1 + _1 = ρ + ⇒ = _0 + _1 + \frac{ρ_0ρ_1}{ρ_0 + ρ_1}(_0 - _1)(_0 - _1),\\
_0 + _1 = ⇒ = _0 + _1.
$$
If, on the other hand, you do not follow a body's contents, by adopting a velocity field whose flow does not go with the mass density, then continuity law for the mass will no longer be homogeneous, but will take the form
$$\frac{∂ρ}{∂t} + ∇·(ρ) = μ.$$
That's equivalent to dividing a body into parts where the enclosing boundary may vary with time - in a way that cuts progressively across the body. Under transform, we have:
$$\begin{align}
\frac{∂ρ}{∂t} + ∇·(ρ)
&→ \left(\frac{∂}{∂t} + ·∇\right)ρ + ∇·(ρ( - ))\\
&= \frac{∂ρ}{∂t} + ∇·(ρ),\\
\frac{∂(ρ)}{∂t} + ∇·(ρ + )
&→\left(\frac{∂}{∂t} + ·∇\right)(ρ( - )) + ∇·(ρ( - )( - ) + )\\
&= \frac{∂(ρ)}{∂t} + ∇·(ρ + ) - \left(\frac{∂ρ}{∂t} + ∇·(ρ) \right),
\end{align}$$
thus
$$μ → μ, \quad → - μ.$$
To make a boost-invariant out of $$ will then require associating the mass flux $μ$ with a momentum flux $$ and an associated velocity $_μ = /μ$. This is the rate at which the mass and momentum are passing through the boundary of the region $Ω$ associated with the body, due to the fact that the velocity field $$ is not following the body's mass flow. Then we can define a boost-invariant version $_0$ of $$ by $_0 = - $, with the transforms:
$$_μ → _μ - , \quad → - μ, \quad _0 → _0.$$
Similarly, for the body, itself, we have the transform:
$$\frac{dm}{dt} → \frac{dm}{dt}, \quad \frac{d}{dt} → \frac{d}{dt} - \frac{dm}{dt}\quad→\quad → - \frac{dm}{dt}.$$
So, associating a momentum flux $$ and velocity $_{\dot{m}}$ with the mass flux $dm/dt$, with
$$ = \frac{dm}{dt}_{\dot{m}}, \quad _{\dot{m}} → _{\dot{m}} - , \quad → - \frac{dm}{dt},$$
we can define the boost-invariant force by:
$$_0 = - .$$
The force law will then read:
$$\frac{d}{dt} = = _0 + = _0 + \frac{dm}{dt}_{\dot{m}}\quad→\quad\frac{d}{dt} - _{\dot{m}}\frac{dm}{dt} = _0.$$
