Second Friedmann (acceleration) equation

Start from the first Friedmann equation in the form $\dot a^2 = \dfrac{8\pi G}{3}\rho a^2 - kc^2 + \dfrac{\Lambda c^2}{3}a^2$ and differentiate both sides with respect to $t$: $2\dot a\ddot a = \dfrac{8\pi G}{3}\left(\dot\rho a^2 + 2\rho a\dot a\right) + \dfrac{2\Lambda c^2}{3}a\dot a$.

Divide every term by $2a\dot a$ (valid whenever $\dot a\neq0$): $\dfrac{\ddot a}{a} = \dfrac{4\pi G}{3}\left(\dfrac{\dot\rho\, a}{\dot a} + 2\rho\right) + \dfrac{\Lambda c^2}{3}$.

Now eliminate $\dot\rho$ using the fluid equation $\dot\rho + 3\dfrac{\dot a}{a}\left(\rho + \dfrac{p}{c^{2}}\right) = 0$, rearranged as $\dfrac{\dot\rho\,a}{\dot a} = -3\left(\rho+\dfrac{p}{c^2}\right)$. Substituting: $\dfrac{\ddot a}{a} = \dfrac{4\pi G}{3}\left(-3\rho-\dfrac{3p}{c^2}+2\rho\right)+\dfrac{\Lambda c^2}{3} = \dfrac{4\pi G}{3}\left(-\rho-\dfrac{3p}{c^2}\right)+\dfrac{\Lambda c^2}{3}$.

This is exactly $\dfrac{\ddot a}{a} = -\dfrac{4\pi G}{3}\left(\rho + \dfrac{3p}{c^{2}}\right) + \dfrac{\Lambda c^{2}}{3}$, as claimed. Notice the fluid equation itself is nothing more than the first law of thermodynamics $d(\rho c^2 a^3) = -p\,d(a^3)$ applied to an expanding comoving volume: energy inside a comoving patch changes only because the pressure does work as the patch's volume grows.