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 (AcOH 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.)

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.)

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 (AcOH 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.)

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 (AcOH 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.)