Newtons laws of motion provide a simple explanation for why drag is proportional to velocity squared of a moving object. Use the example of a falling skydiver that is pulled downwards by gravity; but this explanation also applies to cars, trains, airplanes, etc .... that are being pushed through the air.
The skydiver will fall through a mass of static air ('m') which they push out of their way and accelerate ('a'). This creates a downward force (Force = ma) by the skydiver on the air, according to Newtons 2nd law of motion. The 'equal and opposite' upward force is called drag (Newtons 3rd law of motion). Drag equals the downward force exerted by the falling skydiver to push the air out of their way.
OK. So, if (prior to terminal velocity), the skydiver was to double their downward velocity. Then: (i) The mass of air fallen through each second will double (m x2). (ii) The skydiver will hit each air molecule with twice the momentum as before and thus double the acceleration of each air molecule hit (a x2). The combined effect of theses two is to quadruple the downward force (Force x4 = 2m x 2a). Consequently, the 'equal & opposite' drag on the sky diver will quadruple (Drag x4). Simple. If the skydiver's velocity doubles then drag will quadruple. This explanation isn't in any textbook.
In this process, energy and momentum is transferred from the skydiver to the air. So no net loss or gain of mass, momentum or energy in this process.
The skydiver hits terminal velocity when gravity (the force acting on the skydiver) can no longer accelerate the skydiver to a higher velocity. But don't let gravity or terminal velocity confuse you, they are not critical to explaining the relationship between drag and the skydiver's velocity.
This explanation can be applied to any object falling through a fluid; Such as a stone falling through water or air.