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.
Second, if $E \otimes_k F$ 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 disjoint, namely the fraction field of $E \otimes_k F$. In the latter case, $E$ and $F$ are not linearly disjoint in any compositum.
If $E$ and $F$ are both transcendental extensions, then there are two different criteria: Weakly linearly disjoint, when $E \otimes_k F$ is an integral domain, and strongly linearly disjoint, 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 Finite, connected, semisimple, rigid tensor categories are linear.)
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 disjoint, 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 disjoint extensions have different behavior.
I think that your question is formulated to broad to give a precise answer.
I assume that K is given in the form F(x1,...,xk)/(f1,..,fm). If your equations f_1...f_m are nice enough (low degre etc.) then several computer algebra packages can tell you transcendence degree of K/F using e.g. Groebner Bases. If this tr. degree is different from one then you are done, otherwise K is the function field of a smooth projective curve C/F.
Now the function field of C/F is isomorphic to F(x) if and only if C is isomorphic to ℙ1 .
If F is algebraically closed then it suffices to show that C has genus 0. If char(K)=0 this is can be done relatively easy: following the proof of the lemma of the primitive element you can write K=F(x,y)/f. One can consider y as a function on C and this yields a morphism g:C→ ℙ1. The Riemann-Hurwitz formula (RH) gives you an easy recipe to calculate the genus of C.
(Most of the details are explained in Fulton's book on algebraic curves, if only care about RH you might also read the first two chapters of Silverman's book on Arithmetic of elliptic curves.)
If F is alg. closed, but char(K)>0 this recipe also works, except that the calculation of the entries in the RH formula is harder.
So if F is alg closed then a computer can do the job.
If F is not alg. closed you have a harder problem. You need to use the following criterion
a smooth projective curve C is isomorphic to ℙ1
if and only if
the genus of C is zero and C has a point with coordinates in F.
So besides the genus calculation you need to show that the curve C has at least one point with coordinates in F. This is in general a hard problem, if F is finite this is known and if F is a number field there is a good criterion to check whether C has a point with coordinates or not, for arbitrary F this seems hard.
Examples ℂ[x,y]/(x^2+y^2+1) is isomorphic to ℂ[z], but ℝ[x,y]/(x^2+y^2+1) is not isomorphic to ℝ[z], since the conic x^2+y^2+1 has non ℝ-points.
Best Answer
I think they're equivalent. Clearly 3 ==> 2. To see 2 ==> 3, say an extension E/L of fields is 2-filtered if it has a tower of subextensions each of degree 2 over the next. Then note firstly that if E/L and E'/L are 2-filtered, then so is the compositum of E and E' in any extension of L containing both (induct on the length of the tower of, say, E); and secondly for any map f:E-->E' of extensions of L, if E is 2-filtered then so is f(E). Together these two facts imply that in 2 we may assume that K is Galois over Q, say with group G. Then Q(z)/Q corresponds to a subgroup H, and by Galois theory we just need to prove the following group-theoretic lemma:
Lemma: Let G be a 2-group and H a subgroup. Then there is a chain of subgroups connecting H to G where each is index 2 in the next.
Proof: Recall that we can find a central order two subgroup Z of G. If H\cap Z=1, then HZ gets us one up in the filtration. Otherwise H\cap Z=Z, i.e. Z is in H, and we can use induction on |G|, replacing G by G/Z.