Dual Bundles' Local Trivializations Confusion $$\newcommand{\R}{\mathbb R}$$
Let $\pi:E\to M$ be a rank $k$ smooth vector bundle over a smooth manifold $M$. I will below describe how to form the dual bundle, wherein lies my question.
Let $E_p=\pi^{-1}(p)$ for each $p\in M$, and define $E^*=\bigsqcup_{p\in M}E_p^*$. Further define $\pi^*:E^*\to M$ as $\pi(E^*_p)=\{p\}$ for all $p\in M$.
The main part is to define the local trivializations. For each local trivialization $\Phi:\pi^{-1}(U)\to U\times \R^k$ of $E$ over $U$, define $\Phi^*:\pi^{*-1}(U)\to U\times \R^{k*}$ as
$$\Phi^*(\omega)=(p, (\Phi|_{E_p})^{-t}\omega)$$
for all $\omega\in E_p^*$, and all $p\in U$. (The '$t$' denotes the transpose).
The problem with the above definition is that a local trivialization is supposed to map $\pi^{*-1}(U)$ to $U\times \R^k$ and not $U\times \R^{k*}$.
Now I know that $\R^k$ can be canonical identified with $\R^{k*}$ provided we have an inner product on $\R^k$.
But then different inner products will lead to different $\Phi^*$ and consequently different transition functions.
Is there a preferred inner product with respect to which the identification of $\R^k$ with $\R^{k*}$ is made?

The identification of $V$ with $V^*$ via a scalar product is done for vector spaces in which there is no preferred basis. This is not the case of $\mathbb R ^n$ which has the usual basis (in terms of which it is constructed, after all) that allows you to define the identification $e_i \mapsto e^i$ where $e^i (v) = e^i (\sum v^j e_j) =v^i$. So, the idea is that $\mathbb R^n$ identifies with $(\mathbb R^n)^*$ purely algebraically, with no need for a scalar product.
