Why don't we just say that the Klein Gordon equation describes a two component complex function? These vectors form the basis vectors of the field that the KG equation describes: (for each $\vec{p}$ in $R^3$):
$$|e^{i\vec{p} \cdot \vec{x}} , E=+\sqrt {p^2+m^2}\rangle$$
$$|e^{i\vec{p} \cdot \vec{x}} , E=-\sqrt {p^2+m^2}\rangle$$
Clearly, any single component function $\psi (x)$ can be expressed as a linear combination of only half of these basis vectors.
If we utilise all of these basis vectors in a linear combination, we will end up describing a two component function ($\psi _1 (x), \psi _2 (x)$), not one!
The energy $E$ serves as a label to double the number of basis vectors!! This equation isn't about a single component function.
To put it another way, given a single component initial state $\psi (x)$, you will never know how to evolve it in time using the KG equation! This is because there will be ambiguity about whether to attach $e^{-iEt}$ or $e^{+iEt}$ in the evolution, because each momentum eigenstate is degenerate. This ambiguity goes away with the two component KG equation:
$$\frac {d \psi _1 (x)}{dt}= +\sqrt {P^2+m^2} \psi _1 (x)$$
$$\frac {d \psi _2 (x)}{dt}= -\sqrt {P^2+m^2} \psi _2 (x)$$
 A: You seem confused about what's going on because you do not clearly distinguish between spatial $\vec x$ and total 4-position $x$.
It is true that any function $f(\vec x)$ can be expressed as a "superposition" of either $\mathrm{e}^{\mathrm{i}\vec p\cdot \vec x}$ or its conjugate - that's just the Fourier transform.
It is false that this has anything to do with the Klein-Gordon equation - solutions to the KG equation are functions $f(x)$ on spacetime, and you cannot express such a function as a superposition of $\mathrm{e}^{\mathrm{i}\vec p\cdot \vec x}$s. You could do a 4d Fourier transform and express arbitrary functions on spacetime as "superpositions" of $\mathrm{e}^{\mathrm{i} p\cdot x}$, but note that in this case $p^0$ would be wholly independent from the other components of momentum, since the Fourier transform doesn't know anything about the mass shell.
Instead, the claim about the solution to the Klein-Gordon equation is the following: Any solution $f(x)$ can be expressed as
$$ f(x) = \int \left(A(\vec p)\mathrm{e}^{\mathrm{i}(\vec x\cdot \vec p - tE_p)} + B(\vec p)\mathrm{e}^{-\mathrm{i}(\vec x\cdot \vec p - tE_p)}\right)\frac{1}{2 E_p}\frac{\mathrm{d}^3p}{(2\pi)^3}$$
for arbitrary functions $A(\vec p)$ and $B(\vec p)$. Here we have written the exponentials explicitly in terms of $E_p$, since the shorthand $\mathrm{e}^{\mathrm{i}px}$ seems to be core to the confusion in the question.
This is not a Fourier transform, it really is superposing the specific "basic" solutions $\mathrm{e}^{\mathrm{i}(\vec x\cdot \vec p - tE_p)}$ and $\mathrm{e}^{-\mathrm{i}(\vec x\cdot \vec p - tE_p)}$, and there are lots of functions $f(x)$ that cannot be expressed in this form - if a function has this form, then it is a solution to the Klein-Gordon equation. Additionally, both summands are needed - there is no way to claim we would only need "half" of this.

As for turning the Klein-Gordon equation into a first-order equation:
Your "two-component" solution at the end doesn't really make a lot of sense - what is "$\sqrt{P^2+m^2}$" supposed to be, after all? For an arbitrary function $\psi(x)$, there is no such thing as "$P^2$" - all you can take there is the differential operator $\partial_i\partial^i$, and what even is the square root of a differential operator? You can try to make this work but it's really not something that would be computationally tractable even if you manage to make it well-defined.
Nevertheless, this line of thought ("how do I turn the KG equation into a first-order equation?") is precisely what led Dirac to the Dirac equation. It turns out that in order to have an operator that squares nicely to the KG equation in every component, you need to have four components, not just two, but otherwise, that's exactly what Dirac did - every solution of the Dirac equation (where $\psi$ is a 4-component object whose components are coupled through this equation via the action of the $\gamma$-matrices)
$$ (\mathrm{i}\partial_\mu\gamma^\mu - m) \psi(x) = 0$$
also has the property that every component of $\psi(x)$ individually fulfills the KG equation.
