How to interpret that forces do not depend on acceleration?
An example of a force that depends on position (of the particle) is the force due to a spring:
$$F_x = -kx $$
An example of a force that depends on velocity (of the particle) is the force due to a dashpot
$$F_v = -c\dot x$$
Now, consider a hypothetical force that depended only on the acceleration of a particle:
$$F_a = -d \ddot x$$
The differential equation of motion would then be
$$m\ddot x = -d \ddot x \Rightarrow (m + d)\ddot x= 0$$
Then, unless $m = - d$, the particle's acceleration must be zero.
Next, consider the case that there are position, velocity, and acceleration dependent forces on the particle. The differential equation then becomes
$$(m + d)\ddot x + c\dot x + kx = 0 = m'\ddot x + c\dot x + kx$$
That is, the acceleration dependent force would have the effect of changing the inertial mass of the particle from $m$ to $m' = m + d$. Such a force could be realized by, e.g., an electronic control system.
Typically, a force that does not depend on the particle's position, velocity, etc. but may be time dependent is a driving force and would appear on the right hand side of the equation of motion:
$$m' \ddot x + c\dot x + kx = F_{ex}(t)$$
Finally, the state dependent forces might have time dependence (the system would be time variant).
$$m'(t) \ddot x + c(t)\dot x + k(t)x = F_{ex}(t)$$
$F = ma$ when mass is constant: I think that's a common misconception. It doesn't really represent the law correctly (unless mass is constant).
Newton's second law of motion states, $F = d/dt(mv)$ the external forces acting on an object in an inertial frame is equal to the change in linear momentum (the measure of motion)
I can see why he made that statement: from the equation, the input parameters can be time, mass, velocity (change in position with respect to time).
Because in Newton's second law, you only have the differentials $dx, dv$ and $dt$ (hence why force can only be a function of $x , v$ and $t$. You never have $da$ where $a$ is acceleration (unless Newton's second law was a third order differential equation then you would have $da$). So mathematically you can't really solve Newton's second order differential equation if force was a function of acceleration.