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Depends on what you're referring to as to whether a spring is or isn't a
good model for electric interactions at the atomic level. Note that when
you stretch a wire the wire behaves macroscopically in a spring-like
manner, in that the stretch (strain) is proportional to the applied force
(stress). That means that at the atomic level the distances between
adjacent nuclei lengthen proportional to the applied force, hence the
(electric) forces between neighboring atoms are in fact springlike.
Obviously we're not talking about Coulomb's law here, which refers to two
point charges, we're talking about a complex situation of two neutral
objects containing charged particles interacting with each other.
Another place where this strange representation of electric forces shows
up is in the classical interaction of light with matter, in which the
electrons in atoms are modeled as responding to light as though they were
attached to the atom by a spring (maybe with damping), which sounds very
odd indeed. I remember being very puzzled by this as a student.
Here's a situation that I found instructive. Consider the effect of a
field applied to a hydrogen atom. To first order, the electron cloud is
slightly displaced so that its center is no longer at the location of the
proton, and the hydrogen atom is now an induced dipole. How is the amount
of the electron cloud displacement related to the strength of the applied
field? Make a crude model of the electron cloud as a uniform-density sphere
of charge, with a radius of one Bohr radius, R. Let r represent the
displacement of the proton relative to the center of the electron cloud.
The force that the electron cloud exerts on the proton is due just to that
portion Q of the electron cloud inside the sphere of radius r, where Q =
e(r^3/R^3), so F = ke^2(r^3/R^3)/r^2 = (ke^2/R^3)r, a restoring force
proportional to the displacement r. In equilibrium this force is equal to
the force acting on the proton due to the applied field E, which is F = eE,
so (ke/R^3)r = E, and the displacement of the proton is proportional to the
applied field, which means that you can model the response to an applied
field with a spring-like force.
A footnote to this is that for the polarizability P of the hydrogen atom
we have the dipole moment er = PE = (R^3/k)E, so P = R^3/k, in rough
agreement with experimental measurements. (In cgs units k = 1 and the
polarizability has units of cubic centimeters.)
Bruce