When NaCl is added to water, it dissociates into $\rm{Na^+}$ and $\rm{Cl^-}$. The newly formed $\rm{Na^+}$ and $\rm{Cl^-}$ do not react with water.
When $\rm{CH_3COONa}$ (sodium acetate) is added to water, it dissociates into $\rm{CH_3COO^-}$ and $\rm{Na^+}$. The newly formed $\rm{CH_3COO^-}$ reacts with water to form $\rm{CH_3COOH}$ and $\rm{OH^-}$ ions.
Why is it that $\rm{Cl^-}$ does not react with water, but $\rm{CH_3COO^-}$ reacts with water?
According to Bronsted-Lowry Acid-Base theory, stronger acids have weaker conjugate bases and vice-versa.
$\ce{HCl}$, being a strong acid in water, it's conjugate base i.e $\ce{Cl-}$ is weak or it is more stable. Although it reacts with $\ce{H2O}$ but to a very less extent. Whereas in the case of $\ce{CH3COOH}$, it being a weak acid, it's conjugate base is a strong base. So, it reacts with water and forms relatively more stable $\ce{CH3COOH}$.
The degree of hydrolysis, $\mathrm{h}$ of a salt of weak acid and strong base is approximately given by $\mathrm{h}$=$\sqrt{k_w/Ck_a}$.
If you substitute the $\mathrm{k_a}$ value for $\ce{CH3COOH}$ ( $\mathrm{k_a}$=$\pu{1.8E-5}$) for a $\mathrm{0.1M}$ solution of $\ce{CH3COONa}$, which yields $\mathrm{h=}$ $\pu{7.45E-4}$ where as for $\mathrm{0.1M}$ $\ce{NaCl}$ (although approximate) $\mathrm{h\approx}$$\pu{ 10^-10}$, which is negligible when compared to degree of hydrolysis of $\ce{CH3COONa}$.
