My intuition (maybe misleading?) of the Thom isomorphism is this: just as a vector bundle $\xi:X\to B$ is a twist of the trivial bundle, its Thom space $T\xi$ is a twist of the (rank $\xi$)-fold suspension of $B$ (Thom space of the trivial bundle is just the iterated suspension).

Now for a multiplicative cohomology theory $E$, it seems that this twist manifests itself in the fact that $E^*(T\xi)$ is a rank 1 projective module over $E^*(B)$, i. e. a twist of the free rank 1 module over $E^*(B)$. And picking a $E^*$-orientation of $\xi$ is more or less the same as picking a generator (necessarily of degree rank $\xi$) of this module; in particular, such thing exists iff this module is free, and then the Thom isomorphism is clear - it is just dimension shift by degree of the generator.

Thus one may say that a bundle is $E$-orientable iff $E$ "is not confused by the twist of the iterated suspension introduced by the twist of the trivial bundle caused by $\xi$".

My intuition (maybe misleading?) of the Thom isomorphism is this: just as a vector bundle $\xi:X\to B$ is a twist of the trivial bundle, its Thom space $T\xi$ is a twist of the (rank $\xi$)-fold suspension of $B$ (Thom space of the trivial bundle is just the iterated suspension).

Now for a multiplicative cohomology theory $E$, it seems that this twist manifests itself in the fact that $\tilde E^*(T\xi)$ is a rank 1 projective module over $E^*(B)$, i. e. a twist of the free rank 1 module. And picking a $E^*$-orientation of $\xi$ is more or less the same as picking a generator (necessarily of degree rank $\xi$) of this module; in particular, such thing exists iff this module is free, and then the Thom isomorphism is clear - it is just dimension shift by degree of the generator.

Thus one may say that a bundle is $E$-orientable iff $E$ "is not confused by the twist of the iterated suspension introduced by the twist of the trivial bundle caused by $\xi$".