Why must the Dirac equation multiplied by its complex conjugate give the KG equation? This may be a simple question. I can show this is the case mathematically but cannot explain why it happens. It was only when asked why this happens when I realised I couldn't explain it intuitively/physically. 
So I suppose there must be some way in which the Klein-Gordan equation can be recovered from the Dirac equation, just like a lot of things in Physics. But why does this $\textit{have}$ to be done by multiplying by the complex conjugate and not some other operation? 
 A: The reason is as follows.
The scalar Klein-Gordon equation is (and has to be) a second-order equation because the box
$$\square = \partial_\mu \partial^\mu $$
is the simplest Lorentz-invariant differential operator that may act on scalar fields. However, Dirac wanted to find a first-order equation and it is indeed possible for spinors because
$$ \partial \!\!\!\! / = \gamma^\mu\partial_\mu $$
is also Lorentz-invariant (doesn't change the transformation properties of the object it acts upon), it is a first-order operator, and it is able to act on spinors (spinors are needed because the operator contains the Dirac matrices which must contract their indices with a spinor).
By special relativity, the wave equations must enforce the correct invariant mass. So the de Broglie waves
$$ \exp(ip_\mu x^\mu ) u$$
multiplied by a special form of the spinor (only in the  Dirac case) must be a solution for $p_\mu p^\mu = m^2$. In other words, special relativity implies that a solution to the mass $m$ Dirac equation must be a solution to the mass $m$ Klein-Gordon equation, too.
However, the Dirac equation is stronger because it constrains the first derivatives; the Klein-Gordon equation allows you to choose the wave function and its first derivative(s) and it only constrains the second derivatives.
In the momentum basis, the Dirac equation says
$$ (p\!\!\!/ - m) \Psi = 0$$
which says that the field $\Psi$ must be an eigenstate of "p-slash" with the eigenvalue $+m$. The condition
$$ p\!\!\!/ -m = 0 $$
implies that $p^2=m^2$ because
$$p^2 \equiv p_\mu p^\mu = p\!\!\!/ \cdot p\!\!\!/ $$
due to the anticommutator
$$\{\gamma_\mu,\gamma_\nu\} = 2g_{\mu\nu}\cdot {\bf 1}$$
However, $ p\!\!\!/ =+m $ (when acting on $\Psi$) is not equivalent to $p^2=m^2$ (when acting on $\Psi$). The former condition is stronger, as I said. We may very well have $ p\!\!\!/ =-m   $ and this will still imply $p^2=m^2$ (when acting on $\Psi$). This is nothing else than the claim that there are two solutions to the equation $x^2=m^2$, namely $x=+m$ and $x=-m$.
So the "Klein-Gordon" condition $(p^2-m^2)\Psi=0$ acting on $\Psi$ is simply equivalent to $$(p\!\!\!/ +m) (p\!\!\!/ -m)\Psi = 0$$
which isn't more controversial than the formula $a^2-b^2=(a-b)(a+b)$ and this product may be written as "p-slash plus m" times the Dirac equation.
In other, more physical words, the Klein-Gordon equation allows the positive-energy vector $p^\mu$ as well as $-p^\mu$, its negative-energy opposite. However, $(p\!\!\!/ -m)\Psi=0$, the Dirac equation, only allows one of the signs of the energy and this sign is correlated with the up/down polarization of the electron/positron.
The "other" or "reverted" Dirac equation, $(p\!\!\!/ +m)\Psi=0$, imposes the opposite sign of the energy as the function of the polarization. It's clear that the Klein-Gordon equation (acting on the spinor) which doesn't care about the sign of energy is equivalent to saying that the energy's sign is either what the Dirac equation demands (as a function of the polarization); or the opposite sign. The opposite sign is obtained by the solutions of the "reverted" Dirac equation.
So $\Psi$ obeys the Klein-Gordon equation iff it obeys the Dirac equation; or the reverted Dirac equation; or if it is a combination of these two types of solutions. This is mathematically equivalent to saying that the solution to the Klein-Gordon equation is annihilated by the product of the "Dirac operator" and the "reverted Dirac operator".
The reversal of the Dirac operator is also linked to some kind of complex conjugation, more precisely the charge conjugation (C), because the C-conjugated spinor simply obeys the reverted Dirac equation. This may be checked by defining C more carefully and counting the signs.
