The inside cabin pressure from what I understand is pressurized to the values found at 6000-8000 feet high (11.34 psi at 7000 feet), at which point we can find the amount of force can be found acting against the door internally.
What about the pressures outside of the door? Is it just the pressure of the altitude of the plane acting against the door, or are there other forces involved, such as the force of the wind?
A door that is, say, 36" x 80" (0.9m x 2m), with a surface area of 2880 sq/in (1.8m2), at 8 psi (55kPa) max cabin pressure differential, which most airliners run at, will have roughly 12 tons (102kN) acting on the perimeter fittings trying to blow it out. Any other forces acting on it are microscopic in comparison.
For non plug type doors that depend on bayonet latch pins, the shear loads on the pins are such that it is impossible to move the latching mechanism. If superman was able to move the operating lever, it may or may not just bend the linkage, but no human will be able to move it.
For a plug door of the dimensions above, it has to move inboard to start to open it, so you would have to be able to apply 12 tons of pull force to unplug it.
Just the acrylic cabin windows are holding back over half a ton each when you are cruising along at 38000 ft.
This is why you don't have to worry about a crazy person trying to open a window escape hatch in flight.
If you are talking about passenger doors, their surface will be mostly parallel to the direction of motion, so the wind will not act directly on them (otherwise the aircraft would be in a sideslip, with the drag and fuel consumption greatly increased).
So yes, you can assume only the external pressure is acting on them (plus gravity).
The pressure will generally not be the nominal value, given the fact that the turbulence near the surface of the airplane slightly disturbs it, but neglecting this effect is a good first approximation.
