Let $$\chi_n(z) = \frac{1}{\sqrt{2\pi}}z^{n-1}.$$
Prove $(\chi_n)$ is orthonormal on $\partial B(0,1)$ in regard to $$\langle f(z), g(z)\rangle = \int_{\partial B(0,1)}f(z)\overline{g(z)}\operatorname dS.$$
Choose $\chi_k(z), \chi_j(z)$ then consider:
$$\begin{aligned} \langle \chi_k(z),\chi_j(z)\rangle & = \frac{1}{2\pi}\int_{\partial B(0,1)} z^{k-1} \overline{z^{j-1}}\operatorname d S\\ &= \frac{1}{2\pi}\int_{\partial B(0,1)}z^{k-1}{\overline z}^{j-1} \operatorname d S \end{aligned}$$ New parameterize $z(\theta) = e^{i\theta}$, $\operatorname d z = ie^{i\theta} \operatorname d \theta$ where $\theta: 0 \to 2\pi$, then
$$\begin{aligned} \langle \chi_k(z),\chi_j(z)\rangle & = \frac{i}{2\pi}\int_0^{2\pi}e^{ik\theta-i\theta}e^{-ij\theta+i\theta}e^{i\theta}\operatorname d \theta\\ & = \frac{i}{2\pi}\int_0^{2\pi}e^{i\theta(k-j+1)}\operatorname d \theta \end{aligned}$$
But this doesn't seem right, when $j=k$ this evaluates to zero, where I would want $\langle \chi_k, \chi_j\rangle = \delta_{kj}$
Where am I doing something wrong?
The surface measure $dS$ is not the same as $dz$. In particular, $dS=d \theta$ (here $r=1$), while $dz=i e^{i \theta} d \theta$. Making that correction, your integral is correctly written as
$$\frac{1}{2 \pi} \int_0^{2 \pi} e^{i \theta(k-j)} d \theta.$$
This is readily seen to be $1$ when $k=j$.