Let's understand this statement in Hamiltonian formalism, where KG equation is equivalent to having the free scalar field hamiltonian and the Heisenberg equations of motion for the free fields.
Then $\phi(\vec{x},t) = \int \frac{d^3 p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}\left( a_p e^{ipx} + a^\dagger_p e^{-ipx}\right)$ and the canonical conjugate $\pi(\vec{x},t) = \dot{\phi}(\vec{x},t)$, are the most general solution.
Now let's consider an interacting Hamiltonian $H = H_0 + \lambda V$, and DEFINE $$\Phi(\vec{x},t)\equiv e^{iHt}e^{-iH_0 t} \phi(\vec{x},t) e^{iH_0 t}e^{-iHt}$$
$$\Pi(\vec{x},t)\equiv e^{iHt}e^{-iH_0 t} \pi(\vec{x},t) e^{iH_0 t}e^{-iHt}$$
Then it is straight forward to show that $\Phi$ and $\Pi$ satisfy the canonical commutation relations, as well as the new interacting Heisenberg equations (notice that in the definition we use $H(\phi,\pi)$ and in the Heisenberg equations $H(\Phi,\Pi)$, we are allowed to do so because both are equal!). (Hint: to see that the new fields satisfy the full heisenberg equations notice that $\Phi(\vec{x},t) = e^{iHt}\phi(\vec{x},0)e^{-iHt}$) So in this sense they are the interacting fields, written in terms of the free fields.
Then we conclude that $$\Phi(\vec{x},t) = \int \frac{d^3 p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}\left( \mathbb{a}(t)_p e^{ipx} + \mathbb{a}(t)^\dagger_p e^{-ipx}\right)$$
Where $$\mathbb{a}(t)_p\equiv e^{iHt}e^{-iH_0 t} a_p e^{iH_0 t}e^{-iHt}$$
and $\mathbb{a}_p(t)$ satisfies the required commutation relations as a consequence of its parents $\Phi$ and $\Pi$ doing so, or as can be verified directly using those of $a_p$.
Notice that this description is particularly useful for weakly coupled theories, since then $\mathbb{a}_p = a_p + \mathcal{O}(\lambda,a_p^2)$, then all our particle spectrum can be inferred from that of the free theory, unlike when this expansion is no longer valid, and the new creation operator can create states completely different in nature from what's contained in the free theory.
Nevertheless, in collider experiments, for instance, the particles (or perhaps I should say the quantum fields) clearly are effectively localized into a finite region of space. And there the theory really works!
It works because collider experiments do not measure (x,y,z,t). They measure (p_x,p_y,p_z,E). The calculations are done for point particles entering Feynman diagrams but the numbers that predict measurements are not dependent on space time, but on energy momentum.
No experiment can measure the localization of an individual interaction with the accuracy necessary to see effects of spatial uncertainty: the incoming protons have the Heisenberg uncertainty even if they were measured individually and not as a beam, and the same would be true for the outgoing particles that would have to be extrapolated back to the vertex. Any predictions on the localization of the interaction in the beam crossing region would fall within these combined HUP uncertainties, imo of course.
Best Answer
Your notion of "my previous conception of a particle to be something localized in space" is a classical (non- quantum) conception.
So, in Quantum Mechanics, it it a false conception. And it is the same thing in Quantum Field Theory.
In Quantum Field Theory (as in Quantum Mechanics), you are working with operators. The most natural way - because of the relation with the quantum harmonic oscillator - is to use the impulsion-time representation for these operators : $a(\vec k,t) = a(\vec k) \, e^{i k_0t} , \, a^+(\vec k,t) = a^+(\vec k) \, e^{-i k_0t}$, with $k_0^2 = \vec k^2 + m^2$
But you may use the position representation also :
$$A(\vec x, t) = \int d^4k \,\delta(k^2 - m^2) \, (a(\vec k,t) \, e^{i \vec k.\vec x}+ a^+(\vec k,t) \, e^{-i \vec k.\vec x})$$
$$A(\vec x, t) = \int d^4k \,\delta(k^2 - m^2) \, (a(\vec k) \, e^{i k.x}+ a^+( \vec k) \, e^{-i k.x})$$
So you must be very careful when speaking about "particles". In Quantum field theories, we are working with operators whose eigenvalues are fields (as in Quantum mechanics, the eigenvalues of the Position operator are the only possible measurable positions)
Of corse, instead of speaking about operators, you can speak about states, for instance, the state $|\vec k> = a^+(\vec k)|0>$. But it is a 1-particle state, it is not an operator, and it is not a multi-particle state.
You can also build a "wave-packet" state by considering some particular linear combination of 1-particle states, for instance :
$$ |Wave Packet> = \int d^4 k \,\delta(k^2 - m^2) \, f(\vec k) \, |\vec k>$$
And, if you choose correctly the function $f$, this state could be localized in a precise region of space-time.