I thought that the alkoxide ion would attack the aldehyde to form a secondary alcohol, but apparently, as the two pi-bonds are in conjugation, so, the ion attacks the carbon of the C-C pi-bond. Why is this so?
Your title should read "protic medium" as opposed to "acidic medium". Acidic medium would be EtOH/H+. Your reaction conditions are basic and protic.
Nothing much will happen to the unsaturated aldehyde 1. 1.2-Addition leads to hemiacetal 2. Hemiacetals are unstable to base and reverse to the carbonyl compound. Adduct 3 would eliminate to form 1 in the presence of base. If you employed EtOH-d1 as the solvent, you would incorporate deuterium at the γ-position of 1-d2 via enolate 4. If heptanedial 5 were exposed to EtOK/EtOH, an aldol condensation would ensue to form unsaturated aldehyde 1. Under truly acidic conditions, the diethyl acetal (in the red box) of aldehyde 1 would form. Formation of the β, γ-isomer of 1 is unlikely because the α, β-form 1 is much more stable.