Interpretation of $\phi(x)|0 \rangle$ in interacting theory

All this can be answered by the same object: the Green`s function $\langle 0|\mathcal{T}(\phi(x_1)...\phi(x_n))|0\rangle$ and the polology of it. First, we need to do a fourrier transform:

$$ G(q_1,...,q_n)=\int \frac{d^dq_1}{(2\pi)^d}...\frac{d^dq_n}{(2\pi)^d}\,e^{-iq_1.x_1}...e^{-iq_n.x_n}\langle 0|\mathcal{T}(\phi(x_1)...\phi(x_n))|0\rangle $$

Poles at on-shell $q=q_1+...+q_r$, i.e. $q²=-m²$, indicates the existence of a particle of mass $m$, the existence of a one-particle state $|\vec{p},\sigma\rangle$ in the spectra. The residue of that pole measure the projection of this one-particle state $|\vec{p},\sigma\rangle\langle\vec{p},\sigma|$ with the state: $$ \int \frac{d^dq_1}{(2\pi)^d}...\frac{d^dq_r}{(2\pi)^d}e^{-iq_1.x_1}...e^{-iq_r.x_r}\langle 0 |\mathcal{T}\phi(x_1)...\phi(x_r) $$ and the state $$ \int \frac{d^dq_{r+1}}{(2\pi)^d}...\frac{d^dq_n}{(2\pi)^d}e^{-iq_{r+1}.x_{r+1}}...e^{-iq_n.x_n}\mathcal{T}\phi(x_{r+1})...\phi(x_n)|0\rangle $$ In particular, one can show the LSZ reduction formula by putting each $q_i$ on-shell. You can see all this in the chapter $10$ of the first volume of Weinberg QFT textbook.

Now, if you want to know how to make sense of the state $\phi(x)|0\rangle$ you just need to plug this state into an arbitrary Green's function and work out the residue of various poles. Note that just Lorentz symmetry fixes the projection of this state with one-particle state, up to a overall normalization:

$$ \langle \vec{p},\sigma|\phi_{\alpha}(x)|0\rangle = \sqrt{Z}\,u_{\alpha}(\vec{p},\sigma)e^{-ipx} $$ where this $u$-function is the polarization function, responsible to match the Lorentz indice $\alpha$ with the little group indice $\sigma$. In interacting theories, $\phi_{\alpha}(0)|0\rangle$ is not parallel to $|\vec{p},\sigma\rangle$, since by the polology and LSZ reduction formula above, this would mean that the Green's function of more than two points are zero, implying that the S-matrix are trivial.

Answering both your questions:

1. Yes, the state $\phi_{\alpha}(x)|0\rangle$ correspond to a local preparation, a local field disturbance if you like, because it has the right properties under Lorentz-Poincaré transformations.
2. No, the state $\phi_{\alpha}(x)|0\rangle$ is not a one-particle state because this would imply a trivial S-matrix by the LSZ reduction formula and polology of the Green's function, contradicting the assumption of an interacting theory.