Well your equation balances but that isn't the right reaction.
First lets break this down into half cell reactions.
Oxidation reaction
$\ce{Cu_{(s)} <-> Cu^{2+}_{(aq)} + 2e^{-}}$
Reduction reaction
$\ce{H2O2 + 2 H^+ + 2e^{-} -> 2 H2O_{(l)}}$
Decomposition reaction
Hydrogen peroxide spontaneously decomposes in acid solution.
$\ce{2H2O2 -> 2H2O_{(l)} + O2_{(g)}}$
Stoichiometric reaction between copper and hydrogen peroxide
So the stoichiometric reaction between copper and hydrogen peroxide is:
$\ce{H2O2 + 2 H^+ + Cu_{(s)} -> Cu^{2+}_{(aq)} + 2 H2O_{(l)}}$
the $\ce{2 H^+}$ ions come from the vinegar. (Let's use $\ce{HOAc}$ for vinegar).
$\ce{H2O2 + 2 HOAc + Cu_{(s)} -> Cu^{2+}_{(aq)} + 2OAc^{-} + 2H2O_{(l)}}$
Sodium chloride's Role
Finally the role of the $\ce{NaCl}$ plays in the etch. The $\ce{NaCl}$ must be fairly high to drive the $\ce{Cu^{2+}}$ to a copper chloride complex. This thus reduces the "free" $\ce{Cu^{2+}}$ in solution which keeps the electrochemical potential of copper high so that the solution keeps dissolving copper. (Think of the half-cell copper reaction as if it is in a battery. The voltage of the half-cell would drop as the copper builds up.)