Explanation that air drag is proportional to speed or square speed?

One's naive expectation would be that as the object moves through the medium, it collides with molecules at a rate proportional to $v$. The volume swept out in time $t$ is $A v t$, where $A$ is the cross-sectional area, so the mass with which it collides is $\rho A v t$. The impulse in each collision is proportional to $v$, and therefore the drag force should be proportional to $\rho A v^2$, with a constant of proportionality $C_D$ (the drag coefficient) of order unity.

In reality, this is only true for a certain range of Reynolds numbers, and even in the range of Reynolds numbers for which it's true, the independent-collision picture above is not what really happens. At low Reynolds numbers you get laminar flow and $C_D\propto 1/v$, while at higher Reynolds numbers there's turbulence, and you get $C_D$ roughly constant.

To put it in simple terms, at slow speed the drag is just due to the viscosity of the fluid.

At high speed, the momentum you're imparting to each parcel of air is proportional to the speed, and the number of parcels of air per second you're doing it to is also proportional to speed.

Since force is momentum/second, that's why it's proportional to speed-squared.