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.
Best Answer
It seems your confusion comes from the term "potential". Let me first introduce the Proca equation: \begin{equation} \partial^\mu F_{\mu \nu}+m^2 A_\nu = 0 \end{equation} Where $F$ is the usual Faraday tensor (the gauge curvature tensor). Expanding this equation leads to : \begin{equation} \square A_\nu-\partial_\nu \partial^\mu A_\mu + m^2 A_\nu=0 \end{equation} Then, imposing the Lorenz gauge condition $\partial^\mu A_\mu =0$ leaves us with: \begin{equation} (\square+m^2)A_\mu=0 \end{equation} Well, what have we done here? We considered a spin 1 field, so a field that can be responsible for a force, and we recovered the Klein-Gordon equation for all its components. Since we deal with a spin 1 particle, one can assimilate it to the 4-potential of some force. The time-independent version of this equation is: \begin{equation} (\Delta-m^2)A_\mu=0 \end{equation} When dealing with matter fields, one has a non-vanishing term at the RHS, and to solve the new equation arising from this second term, one use Green functions : \begin{equation} (\Delta-m^2)G(x-y)=-i\delta(x-y) \end{equation} This equation can be solved in the position space as shown in Wikipedia, and leads to the usual Yukawa potential: \begin{equation} G(r)=\frac{1}{4\pi r}e^{-mr} \end{equation} So to conclude, I would say that it is important to keep in mind that the Klein-Gordon equation is not restricted to the spin-0 particles and that the word "potential" in QFT refers to the spin-1 particles.