Full equation when using vinegar, hydrogen peroxide, and salt to etch copper
The acetic Acid is used to dissolve the copper, however it cannot do so directly as the reaction of copper with protons is not spontaneous. 

$$\require{cancel} \ce{Cu_{(s)} + 2H+_{(aq)} \cancel{-&gt;} Cu^2+_{(aq)} + H2_{(g)}}$$
The hydrogen peroxide is used to oxidize the copper since copper does not spontaneously react with acids directly.
$$\ce{\frac12 H2\color{red}{O2} +e- -&gt; OH-}\\ \ce{Cu_{(s)} -&gt; Cu^2+_{(aq)} + 2e-}$$
However, hydrogen peroxide is unstable in acid so the acetic acid is used because it reacts with hydrogen peroxide to form a peroxyacid which is more stable.

$$\ce{CH3C(O)OH + H2\color{red}{O2} -&gt; CH3C(O)\color{red}{OO}H + H2O}$$

This peroxyacid can react directly with copper to dissolve it.

$$\ce{CH3C(O)\color{red}{OO}H_{(aq)} + Cu_{(s)} -&gt; Cu[CH3C(O)O]OH_{(aq)}} \\ \ce{Cu[CH3C(O)O](OH)_{(aq)} + CH3C(O)OH_{(aq)} -&gt; Cu[CH3C(O)O]2_{(aq)} +H2O}$$

As far as the sodium chloride goes, I'm not sure what that is for. My best guess is that it is catalytic in dissolving the copper similar to how salt corrodes steel

$$\ce{CH3C(O)\color{red}{OO}H_{(aq)} + Cu_{(s)} -&gt;[\text{cat.} Cl-_{(aq)}] Cu[CH3C(O)O]OH_{(aq)}}$$

As far as the final species goes, you probably have a complex from the excess chloride

$$\ce{Cu[AcO]2 (aq) + 2NaCl (aq) -&gt; [Cu(AcO)2Cl2]^2- (aq) + 2Na+ (aq)}$$
which is in dynamic equilibrium
$$\ce{[Cu(AcO)2Cl2]^{2-} (aq) &lt;=&gt;[Cl-][AcO-] [Cu(AcO)Cl3]^{2-} (aq) &lt;=&gt;[Cl-][AcO-] [CuCl4]^{2-} (aq)}$$

The color comes from the excess chloride that acts as a ligand to the copper From Wikipedia:


  
  Aqueous solutions of copper(II) chloride. Greenish when high in [$\ce{Cl−}$], more blue when lower in [$\ce{Cl−}$].