You have correctly identified zinc and iron as the anodes, since they are both less noble than copper. In the corrosion reaction, they are both oxidized and dissolved in the aqueous gel.
\begin{aligned}
\ce{Zn ->& Zn^2+ + 2e-} \\
\ce{Fe ->& Fe^2+ + 2e-} \\
\end{aligned}
The electrons flow to the copper cathode, where oxygen, which is dissolved in the liquid medium, is reduced to hydroxide.
$$\ce{O2 +2H2O + 4e- -> 4OH-}$$
As a consequence, the pH of the solution around the copper cathode rises, and the phenolphthalein indicator changes its color from colorless to violet-red in this area.

$\ce{Fe^2+}$ reacts with $\ce{K3[Fe(CN6)]}$ to form a deep blue iron(II,III) hexacyanoferrate(II,III).
$$\ce{K+ + Fe^2+ + [Fe(CN6)]^3- -> KFe[Fe(CN6)]}$$
Therefore, the solution near the iron surface turns blue. Zinc does not react with $\ce{K3[Fe(CN6)]}$ and the solution around the zinc anode remains colorless (source).