Suppose $E_1$ and $E_2$ are two vector bundles over a base $B$ such that $E_1 \oplus E_2 \cong \epsilon$, i.e., the sum is trivial. After taking intersections, we get a cover $U_i$ of $B$ such that $E_1$ and $E_2$ are trivial over the $U_i$, giving transition functions $f_{ij}$ and $g_{ij}$. Thus, possible transition functions for $E_1 \oplus E_2$ are given by $f_{ij} \oplus g_{ij}$. Of course, since $E_1 \oplus E_2$ is trivial, another choice for transition functions is just the identity (although perhaps this also requires information from the isomorphism $E_1 \oplus E_2 \cong \epsilon$).
Given this setup, let's say we know the $f_{ij}$, is there a way to solve for the $g_{ij}$ given that the direct sum is trivial? In particular, there are some examples where I know the structure of $\gamma$ and since $\gamma \oplus \gamma^\perp \cong \epsilon$, I am hoping to compute transition functions for $\gamma^\perp$.
Edit: Per my comment below, $g_{ij}$ may only be determined up to equivalence, but this is all I care about anyway. Moreover, if we assume that $E_2 \cong \epsilon/E_1$ (i.e., assume $B$ is paracompact and let $E_2$ be the orthogonal complement to $E_1$), then $E_2$ is well defined and thus so are the transition maps. By taking quotients, I think it follows that $E_1 \oplus E_2 \cong E_1 \oplus E_1^\perp \implies E_2 \cong E_1^\perp$, assuming $E_1$ gets mapped exactly to itself? Then, probably there is a straightforward way to get transition maps for $E_1^\perp$, although the expression might include the metric. I'll think about this more and may post an answer.