Let $(X, \mathcal{O})$ be a ringed space and $\mathcal{F}$ an $\mathcal O$-module on $X$, which furthermore is assumed to be locally free of some finite rank $n \in \mathbb N$.
Then the dual sheaf $\mathcal{F}^{\vee} := \mathcal{Hom}_{\mathcal O}(\mathcal F, \mathcal O)$ is locally free of rank $n$ again.
Given two open subsets $U_i, U_j \subseteq X$ with intersection $V$ trivializing $\mathcal F$, one obtains a transition function $T_{ij} \in \mathrm{Aut}_{\Gamma(V, \mathcal O)}(\Gamma(V,\mathcal F))$.
It is easy to check that $\mathcal F^\vee$ is trivial on $U_i, U_j$ and hence $V$ as well.
What is the relation between the transition functions $T_{ij}^\vee$ of $\mathcal F^\vee$ and $T_{ij}$ of $\mathcal F$?
It's the transpose inverse.
Let $E$ be a locally free sheaf of rank $n$. Let $i_U$ and $i_V$ trivialize $E$ on open sets $U$ and $V$ respectively. Let $e_i \in O_X(\text{some open set})^n$ have $1$ in the $i$th position and $0$ otherwise.
The isomorphism $O_X^n \simeq \mathcal{Hom}(O_X^n, O_X)$ sends $s = (s_1, ... s_n) \in O_X(U)^n$ to the element that takes dot product with $s$. That is: $(t_1, ... t_n) \rightarrow \sum_{k=1}^n s_k t_k$. The inverse sends $f \in Hom(O_X(U), O_X)$ to $(f(e_1), ... f(e_n)) $, similarly for $V$. I'll denote this by $d_s$.
The trivialization $\mathcal{Hom}(L, O_X) \rightarrow \mathcal{Hom}(O_X^n, O_X)$ is given by $f \rightarrow f \circ i_U^{-1}$ or $f \circ i_V^{-1}$. Call this $j_U$ and $j_V$. The inverses $\mathcal{Hom}(O_X^n, O_X) \rightarrow \mathcal{Hom}(L, O_X)$ is given by $g \rightarrow g \circ i_U$ or $g \circ i_V$.
Now let's put everything together. The matrix representation of the transition matrix on $\mathcal{Hom}(L, O_X)$ is given by $j_V \circ j_U^{-1}$ (which I'll call $T_{ij}^\vee $). Therefore, let $s \in O_X^n(U \cap V)$. We first send it to $d_s$, then it goes to $d_s \circ i_U$. Then, it goes to $d_s \circ i_U \circ i_V^{-1}$.
Note that $i_U \circ i_V^{-1} = T_{ij}^{-1}$. Therefore, we get $d_s \circ T_{ij}^{-1}$. Finally, we evaluate this at $(e_1, ... e_n)$ to find out what $s$ is sent to.
To find the $i$th column of the $T_{ij}^\vee$, we set $s = e_i$, and the $j$th row is the evaluation at $e_j$. $T_{ij}^{-1}(e_j)$ is the $j$th column of $T_{ij}^{-1}$. And then we dot product with $e_i$, so we take the $i$th row.
Therefore, the $i$th column, $j$th row of the matrix for $T_{ij}^\vee$ is the $i$th row, $j$th column of $T_{ij}^{-1}$.