Derivation of Klein Gordon equation from Dirac equation; what does it mean? In Dirac field (Peskin and Schroeder), there is one equation in which it 
multiples the Dirac operator 
$$(-i\gamma^{\mu}\partial_{\mu}-m )$$ 
by
$$(i\gamma^{\nu}\partial_{\nu}-m ),$$
obtaining $\partial^2+m^2 = 0$.
What does it mean?
 A: It means that if a given field is a solution of the Dirac equation, then its components are automatically solutions of the Klein-Gordon equation.
This is basically how the Dirac equation is 'derived' (justified, really). The Klein-Gordon equation is nice and relativistically invariant, but the fact that it's second-order in time is awkward, inconsistent with the form we'd like to have for a Schrödinger equation, and ultimately undesirable. It's therefore desirable to have a "square root" of the Klein-Gordon operator, i.e. something which is first-order in time (and therefore in the form of a Schrödinger equation) and which naturally squares to the Klein-Gordon rendering of the relativistic energy-momentum dispersion relation. This is, of course, impossible using $c$-number coefficients, but having non-commutative operators makes the trick work ─ and of course, when you're finished, you're left with the Dirac equation.
A: All solutions of the free Dirac equation also solve the free Klein-Gordon equation. The reason is that the Klein-Gordon equation is the wave equivalent of Einstein's energy momentum equation, $E^2 = p^2 +m^2 $ obeyed by all free bodies in special relativity. Units such that c=1.
