I have only seen the concept of "abstract linear disjointness of $K_1, K_2$ over $k$" used when at least one of the k-algebras can be embedded in the algebraic closure of the other. (I.e., only in this case can you omit the ambient field.) Both of your examples above are of this form.
In complete generality I don't think the concept of "abstract linear disjointness" makes much sense. For instance, consider two purely transcendental field extensions of $k$, say $k(s)$ and $k(t)$ with s and t algebraically independent indeterminates. (Even this statement seems to be implicitly speaking of some overfield!) The tensor product construction does not see the difference between $k(s) \otimes_k k(t)$ and
$k(t) \otimes_k k(t)$, but inside k(s,t), k(s) and k(t) are linearly disjoint and k(t) and k(t) are not!
There is also a little voice inside my head that says that, when applicable, the right criterion for linear disjointess is that the tensor product be a domain, not a field (as Greg Kuperberg says). Indeed, isn't that what happens in my example $k(s) \otimes_k k(t)$ inside k(s,t) above? But I am going to ignore this little voice and go to bed. We'll see what tomorrow brings.
