TL;DR: Because there is an equilibrium between anion of $\ce{CO3^2-}$ and $\ce{HCO3^-}$ that depletes $\ce{H+}$ and that causes more water to dissociate and form free $\ce{OH-}$.
You should start by reading about hydrolysis and buffers (chemistry). Briefly, you have these equilibria:
\begin{align}
\ce{M+ + OH- &<=> MOH}& &\text{(limiting cases: $\ce{M+}$ = $\ce{Na+}$ or $\ce{NH4+}$)} \\
\ce{H+ + A- &<=> HA}& &\text{(limiting cases: $\ce{A-}$ = $\ce{I-}$ or $\ce{CH3COO-}$)} \\
\ce{H+ + OH- &<=> H2O} &
\end{align}
Here $[\ce{Y}]$ means concentration of $\ce{Y}$. Note that $\frac{[\ce{MOH}]}{[\ce{M+}][\ce{OH-}]} = K_\mathrm{b}$, $\frac{[\ce{HA}]}{[\ce{H+}][\ce{A-}]} = K_\mathrm{a}$ and $[\ce{H+}][\ce{OH-}] = 10^{-14}$.
Depending on the $K_\mathrm{a}$ and $K_\mathrm{b}$ you have several situations. Acids and bases are called strong if they favor dissociation on ions and weak if they favor association of ions.
If both your acid and base is strong, then they both dissociate and the released $\ce{H+}$ and $\ce{OH-}$ are recombining to form water. If acid is strong and a base is weak then acid dissociates fully and yield a lot of $\ce{H+}$, but base doesn't dissociate fully and yield less $\ce{OH-}$. As a result of recombination of $\ce{H+}$ and $\ce{OH-}$, you get excess of $\ce{H+}$ and acidic condition. If the acid is weak and the base is strong you have more free $\ce{OH-}$ and basic reaction much by the same logic. If both acid and base are weak $\mathrm{pH}$ depends on relative strengths on acid and base.
In your case acid is weak ($\ce{CO2}$ forms slightly acidic solution in water) and base is strong ($\ce{NaOH}$ is a strong base). Thus, you have a basic solution.
This story is overly simplified because you have an acid with two stages of dissociation:
\begin{align}
\ce{H2CO3 &<=> H+ + HCO3-} \\
\ce{HCO3- &<=> H+ + CO3^2-}
\end{align}
This adds one equation to the system but similar logic applies.