Proposition 10 tells you that you have $\sum_{g \in G} \chi (g) = 0$ for every character $\chi \in \widehat{G}$ except the principal character.
Then proposition 11 uses proposition 10 and the fact that the characters of $G$ form a group with respect to pointwise multiplication where the inverse of $\chi (g)$ is $\chi^{-1} (g) = \overline{\chi (g)} = \chi (-g)$. In particular, $\chi (g) \overline{\psi (g)} = \varphi(g)$ for some $\varphi \in \widehat{G}$. So by proposition 10 we have $\sum_{g \in G} \chi (g) \overline{\psi (g)} = \sum_{g \in G} \varphi (g) = 0$.
On the other hand, if $\psi = \chi$ we get $\sum_{g \in G} \chi (g) \overline{\chi (g)} = \sum_{g \in G} \chi (g) \chi (-g) = \sum_{g \in G} \chi (g - g) = \sum_{g \in G} \chi (0) = \sum_{g \in G} 1 = |G|$.
Edit(In response to comment)
In this case conjugation corresponds to taking inverses. To see this, note that $\chi : G \to S^1$ because $\chi (g)^{|G|} = \chi (|G|g) = \chi (0) = 1$ hence $\chi (g)$ are roots of unity for all $g$ in $G$ and all $\chi$ in $\widehat{G}$. In particular, $|\chi (g)| = 1 = \chi (g) \cdot \overline{\chi (g)}$ hence $\overline{\chi (g)} = \chi^{-1} (g)$. On the other hand, $1 = \chi (0) = \chi (g - g) = \chi (g) \chi (-g) = \chi (g) \chi^{-1} (g)$ hence $\overline{\chi (g)} = \chi^{-1} (g) = \chi (-g)$.
Hope this helps.