The point is that in relativistic physics the idea of a single-particle wave function as in nonrelativistic wave mechanics does not lead to a sensible quantum theory of relativistic particles. The reason is that collisions of interacting particles at relativistic collision energies lead to production and destruction of other particles, i.e., the relativistic quantum theory should always be formulated as a many-body theory. The best class of models, very successful physics wise (although not understood completely in a strict mathematical sense), are local relaivistic quantum-field theories.
The idea of "canonical quantization" is to make the fields to operators and use the Lagrange formalism to determine the expressions for the quantities as defined by Noether's theorem from Poincare invariance (translations in time and space give energy and momenum, rotations angular momentum, and boosts center-momentum motion).
For the free real Klein-Gordon field the Lagrangian reads
\mathcal{L}=\frac{1}{2} (\partial_{\mu} \phi)(\partial^{\mu} \phi)-\frac{m^2}{2} \phi^2.
Now you write (in this case self-adjoint) field operators for ##\phi##. The Hamilton density is
\mathcal{H}=\Pi \dot{\phi}-\mathcal{L}, \quad \Pi=\frac{\partial \mathcal{L}}{\partial \dot{\phi}}=\dot{\phi}\; \Rightarrow \; \mathcal{H}=\frac{\Pi^2}{2} + \frac{(\vec{\nabla}\Phi)^2}{2} + \frac{m^2}{2} \phi^2.
As you see this is (from a naive point of view) a positive semidefinite self-adjoint operator, and so is the Hamiltonian
H=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \mathcal{H}.
So as a many-body theory the Klein-Gordon equation immediately makes sense.
There's a subtle point here, however. Even in the free case, the Hamiltonian as written above is ill-defined, and a naive mode-decomposition reveals that it's actually divergent. The reason is that we multiply operator-valued distributions at one space-time point, which is not a well-defined procedure. To cure this one introduces normal ordering using the mode decomposition of the free field. Normal ordering means to bring all creation operators to the left of all annihilation operators in this mode decomposition.
Now, the normal-ordering process depends on whether you try to quantize the field as bosons or as fermions. In the fermion case, when you exchange annihilation and creation operators to bring them in normal order, this gives an additional sign, while in the boson case it doesn't. So it makes a difference whether you quantize the KG field as bosons or fermions. It turns out that only the boson choice leads to a positive semidefinite Hamiltonian. That reflects a general theorem: Fields of half-integer spin have to quantized as bosons, those of integer spin as fermions. The KG field has spin 0 and thus must be quantized as bosons in order to have a positive semidefinite Hamiltonian as you want to have to guarantee that the ground state ("vacuum", i.e., no particles present at all) of lowest energy exists. The single-particle energy-momentum eigenstates states describe particles with the usual dispersion relation ##E=+\sqrt{m^2+\vec{p}^2}##. That's why this formalism describes spin-0 bosons.
Sine we've chosen a real field to begin with, we have only one sort of particles here, no antiparticles. This is called a "strictly neutral" particle. The Lagrangian doesn't admit the definition of a conserved charge. In this sense it's "neutral".
You can as well quantize the complex Klein-Gordon field. Then you get particles and antiparticles, and the Lagrangian admits the definition of a conserved. Again you need normal ordering, and you have to quantize the field as bosons. This again leads to a positive semidefinite Hamiltonian and to antiparticles that carry the opposite charge quantum number than particles.
Examples for applications of this formalism are effective models for neutral and charged pions.
