I think now I answer my question.
Although in principle, I can calculate the full character table totally from the orthogonality theorems of characters. Namely, I need to use the two relations:
$$\sum_{k} N_k\chi^{(\Gamma_i)}(C_{k})\chi^{*(\Gamma_j)}(C_k) =\delta_{ij}$$
$$\sum_{\Gamma_j}N_k\chi^{(\Gamma_j)}(C_{k})\chi^{*(\Gamma_j)}(C_{k'})= \delta_{kk'}$$
But in practice, it is not always easy to use these two relations to calculate the characters. Even for $D_3$ group which has only two three irreducible representations, we need proper initial guess if we don't want to solve the set of equations.
On the other hand, if we know the basis functions in the first place, then we can effectively get the characters.
For example, if we want to calculate the characters of $A_1$ of $C_{2v}$ group. First, I need to choose a basis function. Let's say I choose $z$ as my basis function (I have listed the basis functions of $A_1$ in my question), since the irreducible representation is of one dimension, I simple need to see how the operations will change the sign of $z$.
There are four classes for $C_{2v}$ group, and the representatives are $E$, $C_2$, $\sigma_v(xz)$ and $\sigma_{v}^{'}(yz)$
$E$
It is evident that $E$ won't change the sign of $z$, so the character is 1.
$C_2$
The principle axis is $z$, so the character is also 1.
$\sigma_v(xz)$
It is a reflection across $xz$ plane, but it doesn't change the sign of $z$, so the character is still 1.
$\sigma_v^{'}(yz)$
It is a reflection across $yz$ plane, the character is also 1.
So, based on the above discussion of using basis function to calculate characters, we see that basis functions can be readily used for calculating characters. Although the example is really trivial (we can write down the characters of $A_1$ even without thinking), but the example indicates that if we need to calculate the characters of $A_2$, we just don't need to solve the set of equations.
