Work done while compressing an ideal gas (the physical significance of $\int \mathrm dp\,\mathrm dV$)

Let me try to convince you that $ \int dPdv$ is almost negligible. As you have said, $P_{ext} = P_{int} + dP$ but what $dP$ really is? Well I think it is better to assume $dP$ as very small number and hence just adding it to $P_{int}$ will give a value bigger than $P_{int}$ at any moment no matter whatever $P_{int}$ is. So, in this sense $dP$ is just acting as constant. Let's see what this angle of thinking about $dP$ can lead to $$ W_{ext} = \int_{V_i}^{V_f} (P_{int}+dP)dV$$ $$ W_{ext} = \int_{V_i}^{V_f} P_{int}dV + \int_{V_i}^{V_f}dPdV$$ Now, let's just focus on the $dP$ part $$ X= dP\int_{V_i}^{V_f}dV$$ as $dP$ is constant. $$X= dP (V_i - V_f)$$ We agreed that $dP$ is a very small number and hence if we multiply it with any other thing no matter what the result will be very very small and therefore $X$ will be a very small number. $$ W_{ext} = \int_{V_i}^{V_f} P_{int}dV + X$$ Now, we can neglect $X$ and hence write $$ W_{ext} = \int_{V_i}^{V_f} P_{int}dV = W_{int}$$. Your argument that $ \int dPdV$ is negligible is quite sloppy as the integral adds many many pieces of small things ($f(x)dx$ is a very small number as $dx$ is very very small but adding many many of them would produce a different result).
Even in mechanics, when calculate gravitational potential energy we take the working force to be just a little more than $mg$ and hence calculate the work done just plugging the work with $mg$, however, the actual force is more than that.
I said that your argument was sloppy because it’s a matter of hyperreal numbers that when and when we cannot consider something negligible, your argument is vey all right if we just accept the rules of differentials.

...$W_{ext}-W_{int}=\int dPdV$. What does $\int dP dV$ mean physically?

Note that in the "non-approximate" case, we have assumed that $P_{ext}\neq P_{int}$. More precisely $P_{ext}-P_{int}=dP$. Now let's assume that the ideal gas is stored in a container with a movable piston(of a finite mass $m$, but ignore gravity) of area $A$ on top. For now, let's assume that there is no friction. So to do external work, you(or rather, surroundings) are applying a pressure $P_{ext}$(which corresponds to a force $F_1=P_{ext}A$) and the gas is doing internal work by applying a pressure $P_{int}$(which corresponds to a force $F_2=P_{int}A$).

Now let's analyze the forces on the piston. So piston has an upward force of $F_2$(applied by the gas) and a downward force $F_1$ applied by the surroundings. So in this case the net force in the downward direction is,

$$dF_{net}=m(da_{net})=F_1-F_2=P_{ext}A-P_{int}A=(P_{ext}-P_{int})A=dP×A$$

$$\therefore dK = Fds=dP(Ads)=dPdV$$

where $dK$ is the infinitesimal change in the kinetic energy of the piston, and $dV=Ads$ is the infinitesimal change in the volume.

There you have it. You see, there is an infinitesimally small(yet non-zero) net force on the piston which gives an infinitesimally small(yet non-zero) acceleration to the piston. And this infinitesimal acceleration increases the speed of the piston from $0$ to some infinitesimally small velocity. And thus the piston gains an infinitesimal amount of kinetic energy. And the $\int dPdV $ term accounts for this change in kinetic energy.

I know the last paragraph is heavily populated with "infinitesimals", but it is just to show you the insignificance of the motion of the piston. Now what if friction would have been present? In that case, the piston won't move in the first place. But if we also assume that the force due to friction is infinitesimally small, then yeah, the piston would move. But this time it would have a lower value of that infinitesimal acceleration. And, also, it will lose some of its kinetic energy in the form of heat(due to frictional losses).

Summary :- The $\int dPdV$ term accounts for the infinitesimal change in the kinetic energy of the piston.

I hope this is what you meant by "physical interpretation".