What would be an example of characters forming a complete, orthogonal basis for class functions? I understand characters are traces of the matrix of a representation $D(g): G \rightarrow GL(V)$ in representations, but I really fail to understand how these functions can form a basis for the class functions. From how I see it, they are nothing but numbers.
Perhaps someone could shed light on this problem for me by giving an example, maybe on the $S_3$ representation or something, please? Thank you!
 A: So we fix a finite group $G$. Let $D : G \to GL(V)$ be a representation where $V$ is a complex finite dimensional vector space. We notice that $\operatorname{tr}(D(ghg^{-1})) = \operatorname{tr}(D(h))$, so to any representation $D$ on a finite dimensional vector space, we obtain a class functions, that is a function $f : G \to \Bbb C$ so that if $g$ and $g'$ are conjugated then $f(g) = f(g')$. 
Now as you suggested we will work the $S_3$ example.
First, the conjugacy classes in $S_3$ are $e, (12)$ and $(123)$. We write these classes $c_0,c_1$ and $c_2$. A function invariant under conjugaison can be written as $\lambda c_0 + \mu c_1 + \delta c_2$ where the greek letters are complex numbers.
Now, we have three irreducible representations : the trivial representation $\rho_0$, the representation $\rho_1$ which is the sign representation. Finally there is a third representation, given on $\Bbb C^2$ seen as $\{(x,y,z) \in \Bbb C^3 : x + y + z = 0 \}$. 
For example, the trace of $\rho_0$ is just the function equal to $1$, that is $c_0 + c_1 + c_2$. The sign representation has trace $1$ on $e$ and $(123)$ and $-1$ on $(12)$ so the function is $c_0 - c_1 + c_2$. 
As an exercise, you can compute the trace for the third representation and verify that the three functions you obtained are linearly independant. 
Edit : In the vector space of class functions (i.e conjugaison invariant functions $\phi : G \to \Bbb C$), there is a natural product given by $\langle \phi_1, \phi_2 \rangle = \frac{1}{|G|} \sum_{g \in G} \phi_1(g)\overline{\phi_2}(g)$. The orthogonality relations says that if $\chi_1, \chi_2$ are the trace functions of irreducible representations $V_1,V_2$, then $\langle, \chi_1, \chi_2\rangle = 1$ if $V_1 \cong V_2$ and $0$ else. 
For example, $\langle \rho_0, \rho_1 \rangle = \frac{1}{6} (1 - 3 + 2) = 0$ and $\langle \rho_0, \rho_0 \rangle = \frac{1}{6} (1 + 3 + 2) = 1$.  
A: For $S_3$ we have
$$ \begin{array}{l|ccc|ccc}
g\in S_3 & \chi_1(g) & \chi_2(g) & \chi_3(g) & f_1(g) & f_3(g) & f_3(g) \\ \hline
e & 1 & 1 & 2 & 1 & 0 & 0 \\ %\hline
(12) & 1 & -1 & 0 & 0 & 1 & 0 \\
(23) & 1 & -1 & 0 & 0 & 1 & 0 \\
(13) & 1 & -1 & 0 & 0 & 1 & 0 \\ %\hline
(123) & 1 & 1 & -1 & 0 & 0 & 1 \\
(132) & 1 & 1 & -1 & 0 & 0 & 1 
\end{array} $$
which shows the three irreducible characters $\chi_1$, $\chi_2$, $\chi_3$ and three other class functions $f_1$, $f_2$, $f_3$.
It should be clear that $\{f_1,f_2,f_3\}$ are a basis for the space of all class functions, so this space has dimension $3$. One can also check that $\{\chi_1,\chi_2,\chi_3\}$ are linearly independent, and since there are just as many $\chi_i$s as there are $f_i$, the $\chi_i$s must be a basis too.
And you can compute directly that every pair of different $\chi_i$s are orthogonal to each other.
