A possible answer is the following. Assume
$h^0(L)=n, \quad h^0(L \otimes \mu)=m$,
set
$\phi \colon \tilde{C} \to C, \quad \psi \colon C \to \bar{C} \subset \mathbb{P}^{n-1}$
and let $f=\psi \circ \phi \colon \tilde{C} \to \mathbb{P}^{n-1}$ be the composition.
Then $f^*\mathcal{O}_{P^n}(1)=\phi^*L$. On the other hand, by projection formula we have
$h^0(\phi^*L)=h^0(L) + h^0(L \otimes \mu)=n+m$,
and this shows that the complete linear system $|\phi^*L|$ induces a map $g \colon \tilde{C} \to \mathbb{P}^{n+m-1}$. Moreover, the map $f$ is obtained by composing $g$ with the projection from the linear subspace $\mathbb{P}^{m-1} \subset \mathbb{P}^{n+m-1}$ corresponding to the natural inclusion $\phi^* H^0(L \otimes \mu) \subset H^0(\phi^*L)$.
The easiest example is perhaps the following. Assume that $C$ is a genus $2$ curve and take $L=K_C$. Then $\bar{C}=\mathbb{P}^1$, $\tilde{C}$ is a hyperelliptic (this can be proven) genus $3$ curve and $\phi^*L=K_{\tilde{C}}$. We have
$h^0(L)=2, \quad h^0(L \otimes \mu)=1$
and the map $g \colon \tilde{C} \to \mathbb{P}^2$ is precisely the canonical map of $\tilde{C}$. It \tilde{C}$, which is the a double cover of a conic $D \subset \mathbb{P}^2$; the mathbb{P}^2$. The map $f$ f \colon \tilde{C} \to \mathbb{P}^1$ is a quadruple coverof $\mathbb{P}^1$, and is , obtained by composing $g$ with the projection of $D$ from the point in $\mathbb{P}^2$ corresponding to the unique non-zero section of $L \otimes \mu$.

