Let $V$ be a vector space over the field $K$ and $V^*=\mathcal{L}(V,K)$ its dual space. We can prove that $V$ is naturally isomorphic to its double dual $V^{**}$, but why does every isomorphism between $V$ and its dual $V^*$ depend on the choice of basis? We certainly use dual basis, but the number of element of a basis i.e. the dimension is not basis-dependent. For example, in Linear Algebra by Serge Lang, I have found this:
Let $V$ be a vector space over $K$ with a non-degenerate scalar product, $\langle\cdot,\cdot\rangle:V\times V\rightarrow K$. Let $v\in V$, the map $L_v$ such that \begin{equation} V\ni u\overset{L_v}{\longrightarrow}\langle u,v\rangle \end{equation} is a linear functional, thus an element of $V^*$.
The map such that \begin{equation} V\ni v\rightarrow L_v \end{equation} is an isomorphism (between $V$ and its dual). This is proved by showing that this map is linear, injective (because of non-degeneracy) and surjective ($dimV=dimV^*$). So, how does this depend on the choice of basis? It is true that we used the dual basis at the beginning, but as I said above, every basis would give us the same answer as for the dimension of the space, that is what the author used in the last proof.
It's not natural because the identification between elements of the vector space and its dual depend upon the basis and, importantly, the correspondence is not maintained upon a change of basis. The latter is easiest to see by thinking about how the components of vectors and convectors change in opposite senses under a change of basis. In your case, note that your map defines dual vectors with respect to the basis dual to that of the original space which is a basis dependent identification.
This is not the case for the correspondence between the dual of the dual and the original space, where a basis-independent identification can be made. In this way, although (for finite dim vector spaces) any vector space of the same dimension is equivalent to the original space, the identification $(V^*)^* \cong V$ is natural because everyone will agree on the correspondence between vectors regardless of the basis they're using.