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Let $\phi^{*}:W^* \rightarrow V^*$, which maps linear function $\alpha \in W^*$ into linear function $\phi^{*}(\alpha)=\alpha \circ \phi$, where $\phi: V \rightarrow W$.

Now let $\mathbb{e}, \mathbb{f}$ are bases of $V$ and $W$, $\mathbb{e}^*, \mathbb{f}^*$ are dual bases to them. We know that linear map $\phi$ has matrix A in bases $\mathbb{e}, \mathbb{f}$. So, what is matrix of $\phi^{*}$ in bases $\mathbb{e}^*, \mathbb{f}^*$?

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    The transposed matrix ${}^{\mathrm t\!}A$. – 2017-02-20
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    Be careful, it is $A^\top$ in bases ${\bf f}^\ast$ and ${\bf e}^\ast$. Order matters :) – 2017-02-20

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Be careful with the order of the bases. If $A = [\phi]_{{\bf e},{\bf f}}$, then the only thing that makes sense to investigate is $[\phi^\ast]_{{\bf f}^\ast, {\bf e}^\ast}$. We'll show that this last matrix equals the transpose $A^\top$. You know that $\phi(e_j) = \sum_i a_{ij}f_i$. We want to show that $\phi^\ast(f_j^\ast) = \sum_i a_{ji}e_i^\ast$. We have $\phi^\ast(f_j^\ast) = f_j^\ast \circ \phi$. Let's evaluate both sides at some $e_k$:$$\phi^\ast(f_j^\ast)(e_k) = f_j^\ast \circ \phi(e_k) = f_j^\ast \left(\sum_i a_{ik}f_i\right) = \sum_i a_{ik}f_j^\ast(f_i) = \sum_ia_{ik}\delta_{ji}=a_{jk}.$$On the other hand: $$\sum_i a_{ji}e_i^\ast(e_k) = \sum_i a_{ji}\delta_{ik} = a_{jk}.$$So they're equal.