How precisely the Klein-Gordon equation is derived? In various articles (and books) such as the wiki article of the Klein-Gordon equation wrote: 
"The Klein-Gordon equation is a "quantized" version of the relativistic energy-momentum relation";
In the article of "canonical quantization" has been written:
"The Klein-Gordon equation is the classical equation of motion for a free massive scalar field, but also the "quantum" equation for a scalar massive particle wave-function.";
I'd like to know,
1) What is exact name of the procedure of derivation of Klein-Gordon equation from the energy-momentum relation? (based on the above articles, i.e. by replacing quantities energy and momentum and so on with their corresponding quantum operators)?
2) Finally, the Klein-Gordon equation is a classical massive field equation, or a quantum mechanical massive wave-function equation, or both?
This question may sound elementary, however, I would appreciate if somebody clearly and simply answer it.
 A: In Quantum Mechanics, the Schrödinger equation is just the statement that energy is the generator of time evolution. In the QM framework this is written as
$$H|\psi(t)\rangle=i\hbar\dfrac{d|\psi(t)\rangle}{dt}.$$
Now, if we have the position representation $\mathbf{r}$ we can form the wavefunction $\Psi(\mathbf{r},t)=\langle \mathbf{r}|\psi(t)\rangle$ and this becomes
$$\langle \mathbf{r}|H|\psi(t)\rangle=i\hbar \dfrac{\partial\Psi}{\partial t}.$$
The usual Schrödinger equation is found when we replace $H$ by the quantized classical hamiltonian:
$$H=\dfrac{P^2}{2m}+V.$$
The question is that the equation you get for $\Psi(\mathbf{r},t)$ is not Lorentz invariant. And indeed, we used the non relativistic energy when we quantized.
Now, the canonical way to do it, is to try quantizing the relativistic version
$$E^2=p^2+m^2,$$
in units where $c=1$. To quantize this we insist that energy is the generator of time translations. This suggests that $E\mapsto i\hbar \partial_t$ while we insist that $p$ is the generator of spatial translations so that $p\mapsto -i\hbar \nabla$. This leads to
$$-\hbar^2\dfrac{\partial^2\Psi}{\partial t^2}=-\hbar^2\nabla^2\Psi+m^2\Psi,$$
or also choosing units where $\hbar =1$
$$(\square+m^2)\Psi=0.$$
Here, $\Psi$ is a wave function, hence $\Psi:\mathbb{R}^3\times \mathbb{R}\to \mathbb{C}$ and hence, despite this strange terminology, $\Psi$ is a classical field.
So for $(1)$, we just quantized the energy momentum relation, by requiring that the same relation holds in the quantum version and imposing that energy is the generator of time translations and momentum the generator of spatial translations.
Now for $(2)$, the Klein-Gordon is a wave function equation. You are just rewriting Schrödinger's equation with a particular Hamiltonian. In the same way, it is a classical field. It is a classical field because it is not operator valued. A quantum field is one operator valued field. Now, talking about making it into a quantum field, that is, dealing with the quantization of this field is another story.
A: The starting point is the representation theory of the Poincaré group (or actually the induced representation theory, and in particular the little group method of Wigner).
For a massive particle of spin zero and mass $m$, the spectrum of the momentum operator is the hyperboloid $p^2 = m^2$, with the energy condition $p^0 > 0$, sometimes denoted by $\Omega_m^+$. One of the advantages of this description is that one gains a genuine invariant measure rather than a just quasi-invariant one, on $\Omega_m^+$, given by
$$\text d\Omega_m^+(p) = \delta(p^2-m^2)\theta(p^0)\text d^4p,\qquad\forall p\in\Omega_m^+.$$
The physical Hilbert space is then $H=L^2(\Omega_m^+,\text d\Omega_m^+)$ and clearly any element of this space satisfies the equation
$$(p^2 - m^2)\phi(p) = 0,\qquad\forall\phi\in H.$$
The Fourier transform of this equation gives the Klein-Gordon equation
$$(\square + m^2)\psi(x) = 0.$$
A: Strictly speaking, the Klein-Gordon equation is not a relativistic version of the Schrödinger equation.
The stationary Klein-Gordon equation is obtained by replacing the relativistic pulse with the momentum operator $p_{relativistic} \mapsto -i\hbar\triangledown$ in the expression for the coupling of the energy and momentum of the STR.
The Klein-Gordon equation has many drawbacks. For example, the erroneous value of the critical charge of the nucleus is Z = 68.
If we act in another way and in the expression for the connection of the energy and momentum of the SRT, we replace the usual pulse  by the momentum operator $p_{Nonrelativistic} \mapsto -i\hbar\triangledown$, then we obtain a completely different equation M2. Equation M2 does not have these drawbacks.
More details on the derivation of the M2 equation can be found in the publication:  http://vixra.org/abs/1609.0086
