The atmosphere exerts a pressure on the upper side of the water surface and that is taken into account in all physics calculations, however molecules present in a liquid move in all directions, and as a result strike the water surface from below, which basically means they are exerting force on the water surface. Since pressure is equal to force per unit area, the molecules are basically applying pressure on the water surface, but from below (opposite to the direction of pressure applied by he atmosphere). Why isn't this pressure taken into account during physics calculations?
Your argument is quite correct. It's just that you have misinterpreted what is meant by the statement that the pressure on the liquid is atmospheric pressure.
The atmosphere exerts a downwards force on the air/water interface, and the water exerts an upwards force on water/air interface. Since the interface is stationary we know the two forces must be equal and opposite, and since they act on the same area the two pressures must be equal and opposite.
So the net force on the interface is zero, but it's zero because the pressure the atmosphere exerts on the water is equal to the pressure the water exerts on the atmosphere. Both pressures are real and non-zero. For example if you measured the density of the water accurately enough you'd find it has been slightly compressed by the pressure the atmosphere exerts on it.
You first have to understand the definition of pressure in a fluid. If we consider a surface inside the bulk of fluid, it will experience equal forces from both sides of the surface. The pressure is defined as the magnitude of one of the forces, acting on unit area. Now at the open surface of the fluid which is exposed to atmosphere, atmospheric force p₀A acts on the surface from above. An equal force acts on the surface from below (since surface is in equilibrium). So the pressure at the surace is the magnitude of one of these forces per unit area, that is p₀A/A = p₀.