A multicomponent signal can be expressed in exponentials can be expressed as follow, $$x(t)=\sum_{k}c_k e^{2\pi f_0 t}$$
The Fourier transform of $x(t)$ is easy $$\mathcal{F}_x(f)=\sum_k c_k \delta(f-kf_0)$$ while the Wigner Function is
$$ W_x(t,f) = \frac{1}{2\pi}\int x\Big(t+\frac{\tau}{2}\Big)x^*\Big(t-\frac{\tau}{2}\Big) e^{-j2\pi f\tau}\;d\tau $$ $$=\sum_k |c_k|^2 \delta(f-kf_0)+\sum_k \sum_{m \neq k} c_k c_m^* \delta(f-\frac{k+m}{2})e^{j2\pi(k-m)f_0t}.$$
However reference for this Wigner function did not provide the derivation and I'm struggling how to properly derive it.
I have already derived the Wigner function of a signal with trigonometric function $x(t)=cos(2\pi t +\theta)$ in this question but I don't know where to start with this multicomponent signal. Any hint will be appreciated.
Hint 1: You seem to have several typos in your provided formulas above. Please read them carefully from the source and fix the transcriptions.
Hint 2: Realize that the term out front, in the expression for the specific Wigner function, is just a special case for when $k=m$, that has been pulled out of the double summation:
$$\begin{align} &\sum_k |c_k|^2 \delta\left(f-kf_0\right)+\sum_k \sum_{m \neq k} c_k c_m^* \delta\left(f-\frac{k+m}{2}f_0\right)e^{j2\pi(k-m)f_0t} \\ \\ &= \sum_k \sum_{m} c_k c_m^* \delta\left(f-\frac{k+m}{2}f_0\right)e^{j2\pi(k-m)f_0t}\\ \end{align}$$
Hint 3: Realize that the product of two summations is going to result in a double summation:
$$\begin{align} x\left(t+\dfrac{\tau}{2}\right)x^*\left(t-\dfrac{\tau}{2}\right) &= \left(\sum_{k=L}^N c_k e^{j2\pi kf_0\left(t+\frac{\tau}{2}\right)}\right)\left(\sum_{m=L}^N c^*_m e^{-j2\pi mf_0\left(t-\frac{\tau}{2}\right)}\right)\\ \\ &=\left(\sum_{k=L}^N c_ke^{j2\pi kf_0t} e^{j2\pi \frac{k}{2}f_0\tau}\right)\left(\sum_{m=L}^N c^*_m e^{-j2\pi mf_0t}e^{j2\pi \frac{m}{2}f_0\tau}\right) \\ \\ &= c_Lc^*_Le^{j2\pi (L-L)f_0t} e^{j2\pi \frac{L+L}{2}f_0\tau }\\ &+ c_Lc^*_{L+1}e^{j2\pi (L-[L+1])f_0t} e^{j2\pi \frac{L+L+1}{2}f_0\tau }\\ &\dots \\ &+ c_Nc^*_{N-1}e^{j2\pi (N-[N-1])f_0t} e^{j2\pi \frac{N+[N-1]}{2}f_0\tau }\\ &+ c_Nc^*_{N}e^{j2\pi (N-N)f_0t} e^{j2\pi \frac{N+N}{2}f_0\tau }\\ \\ &= \sum_k \sum_m c_kc^*_me^{j2\pi (k-m)f_0t}e^{j2\pi \frac{k+m}{2}f_0\tau} \\ \end{align}$$
Getting from that double summation to the specific Wigner function you desire should be very straight forward.