In acidic solution, what does ClO₃⁻ reduce into?
The first equation is correct because $\ce{ClO3-}$ is reduced to $\ce{Cl-}$ and oxygen maintains its -II oxidation state in both educts and products. As chlorine changes its oxidation state from +V in chlorate to -I in chloride, $6 \ \mathrm{e^-}$ are required for this reduction.

In your second equation, chlorate is reduced to chloride as well; however, only $2 \ \mathrm{e^-}$ are explicitly supplied. The remaining electrons would have to come from the concurrent oxidation of oxygen from -II in $\ce{ClO3-}$ to 0 in $\ce{O2}$, and this is unlikely to happen, as oxygen is more electronegative than chlorine.