The set of isomorphism classes of $1$-dimensional vector bundles over $B$ forms an abelian group with respect to the tensor product operation. How do I see that a given $\mathbb{R}^1$-bundle $\xi$ possesses a Euclidean metric if and only if $\xi$ represents an element of order $\le2$ in this group?
A Euclidean vector bundle is a real vector bundle $\xi$ together with a continuous function$$\mu: E(\xi) \to \mathbb{R}$$such that the restriction of $\mu$ to each fiber of $\xi$ is positive definite and quadratic. The function $\mu$ itself will be called a Euclidean metric on the vector bundle $\xi$.
As stated in the comment by @MikeMiller, this can be seen in terms of cocycles. Alternatively, you can observe that both the conditions you consider are equivalent to $\xi\cong\xi^*$, where $\xi^*$ is the bundle dual to $\xi$.
Since $\mu$ defines an inner product on each fiber of $\xi$, it gives rise to an isomorphism $\xi\cong\xi^*$. Now $\xi\otimes\xi^*$ is isomorphic to $L(\xi,\xi)$ and hence canonically trivial. Hence $\xi\cong\xi^*$ implies that $\xi\otimes\xi$ is trivial, so $\xi$ is an element of order two.
Finally, if $\xi\otimes\xi$ is trivial, the trivialization itself defines a quadratic form on each fiber of $\xi$. Taking the negative of the trivialization (on some connected components) if necessary, you can make this positive definite, thus defining $\mu$.