Regard all the fields as subfields of the algebraic closure of K (or K(X)). More precisely, choose an algebraic closure of K and form $k^{1/p}$ inside of it.

Regard all the fields as subfields of the algebraic closure of K (or K(X)). More precisely, choose an algebraic closure of K and form $k^{1/p^{\infty}}$ inside of it.

Added: Linear disjointness is only defined for subfields of some big field. If you choose a different algebraic closure of K, then an isomorphism from it to the first will carry $k^{1/p^{\infty}}$ onto $k^{1/p^{\infty}}$ (with my definition), so you get an isomorphic situation. A similar remark applies to the second example (take $\bar k$ to be the algebraic closure of k in an algebraic closure of K(X)). This is why some authors don't bother to make this explicit (judging by this discussion, they should).

There is no ambiguity.