Why should an elementary particle be a localized perturbation of that particle's quantum field? I'd rather define it as a free one-particle Fock state, which does not need to be localized in any sense of the word. For massless particles with spin ##\geq 1## it's even hard to make sense of "localization" since there's no position operator in the usual sense for them.
The spin of a particle is given by the transformation properties of the zero-momentum states of that particle. For massive particles the rotation subgroup of the Lorentz group leaves a particle at rest at rest, and thus for a massive particle the zero-momentum states allow for a representation of the full rotation group (or more precisely its covering group SU(2), which comes into the game, because not the Hilbert space vectors are the true representants of the (pure) states but rays in Hilbert space, i.e., just multiplying all vectors with the same phase factor doesn't change the state, and thus also unitary representations up to phase factors (socalled ray representations) are allowed, i.e., you are allowed to look at all representations of the group SU(2) to represent rotations). Since the particle is assumed to be elementary this should be an irreducible representation, and thus you can just apply the representation theory of angular momentum from your quantum mechanics course. The representations of SU(2) are completely determined by the spin quantum number ##s \in \{0,1/2,1,\ldots \}##. This means each elementary particle is determined by its mass squared ##m^2## and spin ##s##. For given ##s## a basis are the eigenstates of the z-component of angular momentum, ##\sigma_z \in \{-s,-s+1,\ldots,s-1,s \}##, i.e., for ##\vec{p}=0## you have ##(2s+1)## orthogonal basis vectors.
This is enough to extend the Hilbert space to allow for representations of the full Poincare group, and via boosts, also represented by unitary transformations, you can get any non-zero momentum eigenstate too. So also for any non-zero momentum you have ##(2s+1)## orthogonal independent basis vectors, and this explains why you have these additional descrete spin-degrees of freedom.
