Here is an elementary argument. By Weinstein's tubular neighbourhood theorem we can arrange that locally $L_1$ is the zero section of $T^* \mathbb{R}^n$, and $L_2$ is another Lagrangian intersecting $L_1$ transversely at $0$ which is the image of some other section $\eta(q) : \mathbb{R}^n \to T^* \mathbb{R}^n$. This image is Lagrangian exactly when $d \eta = 0$ (thinking of $\eta$ as a 1-form). Trivializing $T^* \mathbb{R}^n = \mathbb{R}^n \times \mathbb{R}^n$ with coordinates $(q, p)$, we can also write $L_2$ as the graph of the "inverse function" $f(p) = (f_1(p), \ldots, f_n(p))$ of the fiber coordinate $p$.
By the Poincaré lemma we can therefore find a function $H(p) : \mathbb{R}^n \to \mathbb{R}$ so that $\frac{\partial H}{\partial p_i} = -f_i(p)$. Viewing $H$ as a function on $T^* \mathbb{R}^n$ the corresponding Hamiltonian vector field is $$ X_H = - \frac{\partial H}{\partial p_i} \frac{\partial}{\partial q_i} = - f_i(p) \frac{\partial}{\partial q_i}. $$ The time-1 flow of $L_2$ under the Hamiltonian isotopy generated by $X_H$ thus defines a symplectomorphism which fixes $L_1$ and takes $L_2$ to the fiber of $T^* \mathbb{R}^n$ over zero. Identifying my $T^* \mathbb{R}^n$ with your $\mathbb{C}^n$ gives the desired chart.