You're right that in general, the right way to think about electromagnetic interactions isn't between charges at all: instead charges each individually act on the field, which intervenes between them. Newton's third law and its strong form just boil down to overall conservation of linear and angular momentum for both the charges and field together.
However, in most situations where we talk about introductory mechanics, the change in (angular) momentum of the field is negligible. This generally holds as long as particles are not accelerating significantly, and are moving slowly compared to the speed of light.
This can be established heuristically in a few cases. For example, consider two charged particles separated by a distance $r$, with charge $q$ and speed $v$, neglecting emission of radiation. The ordinary electrostatic force between them, which does obey the strong form of Newton's third law, is
$$F_e \sim q E \sim \frac{q^2}{\epsilon_0 r^2}.$$
Meanwhile, the magnetic force between them, which does not obey Newton's third law, is
$$F_m \sim q v B \sim q v \left(\frac{\mu_0 q v}{r^2}\right) \sim \frac{\mu_0 q^2 v^2}{r^2}.$$
The ratio of these forces is
$$\frac{F_m}{F_e} \sim \mu_o \epsilon_0 v^2 \sim \frac{v^2}{c^2}$$
which is indeed small when the charges move nonrelativistically. (Incidentally, the same analysis holds for particles interacting gravitationally, via gravitoelectromagnetism.). To check this, we can also estimate the field momentum. The field momentum density is
$$\mathcal{P} \sim \frac{1}{c^2} \frac{E B}{\mu_0}.$$
The right $E$ and $B$ to use here are the electric field of one particle and the magnetic field of the other. (Taking the same fields for both particles would just give the momentum carried by a particle in isolation, which can be absorbed into the definition of the particle's mass.) The product $E B$ is hence non-singular and significant over a volume of order $r^3$, giving an electromagnetic field momentum
$$P_{\text{em}} \sim r^3 \mathcal{P} \sim r^3 \, \frac{1}{\mu_0 c^2} \frac{q}{\epsilon_0 r^2} \frac{\mu_0 q v}{r^2} \sim \frac{\mu_0 q^2 v}{r}.$$
What matters is the rate of change of this momentum, which is
$$\frac{dP_{\text{em}}}{dt} \sim \frac{\mu_0 q^2 v^2}{r^2}$$
which is precisely the order of $F_m$, i.e. the violation of Newton's third law. So everything checks out; the field picks up the "missing" momentum.
That is precisely why Newton's third law gets mentioned less and less as one continues in the physics curriculum. It's ultimately just an approximation, that ends up replaced with the deeper ideas of momentum and angular momentum conservation.
