As is always the case, the equation of motion is obtained by the variation of the action; for the case at hand, Box φ + V'(φ) = 0, where Box is the d'Alembertian operator. The quadratic part of V is the mass term and the non-quadratic the interaction. The reason is that the most general action, consistent with Lorentz invariance, and containing two derivatives of the scalar field, is the spacetime integral of -(1/2)φ(Box φ) -V(φ), where V(.) is a Lorentz scalar function of φ(x) (up to a total derivative term). The Klein-Gordon equation is the equation of the field, whose excitations are particles.
Lol. The usual procedure in quantum mechanics is to replace derivatives with derivative -eA, where A is a component of the 4-vector defining the external field.