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Let $E$ and $\tilde{E}$ be vector bundles of rank $k$ over $M$, and assume they can be trivialized over the same cover $\{U_\alpha\}$. Suppose $E$ is orientable. The definition of orientable that I know is that the transition maps of $E$, say $t_{\alpha\beta}: U_\beta \cap U_\alpha \to GL(k, \mathbf{R})$ always has positive determinant. It's unclear to me why $\tilde{E}$ must also be orientable if there is a vector bundle isomorphism $\varphi: E \to \tilde{E}$. For example such a vector bundle isomorphism is given by the data of smooth maps $h_\alpha: U_\alpha \to GL(k, \mathbf{R})$ such that $\tilde{t}_{\alpha\beta} = h_\alpha t_{\alpha\beta} h_\beta^{-1}$. In fact, I think you might as well take the $h_\alpha$'s to be arbitrary and define $\tilde{E}$ to be the vector bundle over $M$ with transition maps given by the above formula. But in general it doesn't seem like there's any guarantee that $\tilde{t}_{\alpha\beta}$ has positive determinant, as $h_\alpha$ and $h_\beta$ might have determinants of different signs. Even if you can fix this locally by flipping a sign in a chart, it isn't clear to me that this can be done without causing problems in intersections with other charts.

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Your definition of "orientable" is wrong. Indeed, it may be that $t_{\alpha\beta}$ always has positive determinant for one choice of local trivializations, but not for another.

The correct definition along these lines is rather: a vector bundle $E$ over $M$ is orientable if there exists an open cover $(U_\alpha)$ of $M$ and trivializations of $E$ over each $U_\alpha$ such that the associated transition maps $t_{\alpha\beta}$ always have positive determinant. This definition is obviously preserved by isomorphisms, since it doesn't depend on any specific choice of local trivialization (the local trivialization used is bound by an existential quantifier).

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