The reasonable meaning following example (1) seems to be that $E \otimes_k F$ is a field. If so, then it is isomorphic to every compositum. If not, then there exists a compositum within which they are not linearly disjoint.
I am not (yet) getting voter support, but I stand my ground! :-)
First, clearly if
$E \otimes_k F$ is a field, then it is isomorphic to every compositum.
$E \otimes_k F$ is not a field, then there exists a compositum in which $E$ and $F$ are not linearly disjoint. It has a non-trivial quotient field, and that field can serve as a compositum. As Pete Clark points out, there is a difference between the case that
$E \otimes_k F$ is an integral domain and the case that it has zero divisors. (And Pete is right that I forgot about this distinction.) In the former case, there exists a compositum in which they are linearly independentdisjoint, namely the fraction field of
$E \otimes_k F$. In the latter case, $E$ and $F$ are not linearly independent disjoint in any compositum.
If $E$ and $F$ are both transcendental extensions, then there are two different criteria: Weakly linearly independentdisjoint, when
$E \otimes_k F$ is an integral domain, and strongly linearly independentdisjoint, when it is a field. Which you think is the more important condition is up to you. In Andrew's examples, $E$ and $F$ aren't both transcendental, so the distinction is moot.
(I needed to think about this issue in this paper.)
Actually, the previous isn't the whole story. If $E$ and $F$ are both transcendental, then they are extensions of purely transcendental extensions $E'$ and $F'$. $E'$ and $F'$ are only weakly linearly independentdisjoint, and therefore $E$ and $F$ are too. So the distinction is always moot. Pete and Andrew's intuition was more correct all along. The correct statement is that when $E$ and $F$ are both transcendental, linearly independent disjoint extensions have different behavior.