Let $E/k$, $F/k$ be two arbitrary field extensions of $k$. My question is:
Is there a field extension $M/k$ s.t. $E/k$, $F/k$ are subextensions of $M/k$? Alternatively, can we talk about compositum fields without assuming a larger field?
If the answer to the above question is yes, can we construct such a $M/k$ explicitly (by tensor product, direct product, localization, quotient etc.)?
Is the $k-$algebra $E\otimes_k F $ never the zero ring?
As you seem to have realized the concept of a compositum is a bit troubling if we are not working inside a bigger field. Anyway, here come a few quick and dirty answers:
1+2) Yes (up to identification). The tensor product $E\otimes_k F$ is a commutative $k$-algebra. If $I$ is a maximal ideal in there (call upon Zorn's lemma to get the existence), then $M=E\otimes_k F/I$ is a field. The mappings $e\mapsto e\otimes 1 +I$ and $f\mapsto 1\otimes f +I$ are then homomorphisms of $k$-algebras, and thus injective (scores of details to check here). Therefore we can identify their images with $E$ and $F$ respectively. However, the choice of $I$ may make quite a difference. For example, if $E$ and $F$ are isomorphic (think: $\mathbf{Q}(\root 3\of 2)$ and $\mathbf{Q}(\omega\root3\of2)$, $\omega=(-1+i\sqrt3)/2$), then their images are equal for an appropriate choice of $I$, but may intersect trivially for another one.
[Edit: It is possible that the images of $E$ and $F$ never intersect trivially. A trivial intersection occurs in my example case, but clearly not always. Sorry about any possible confusion this error in the original version may have created.]
3) No. This never happens. The tensor product of two non-trivial vector spaces is never zero. You get a basis for the tensor product from pairwise elementary tensors of basis elements of the factor spaces. When we mod out a maximal ideal as above, some linear dependencies may or may not be introduced (the ideal may be zero).
Consider the tensor product $A=E\otimes _kF$.
a) The $k$-algebra $A$ is nonzero because any choice of bases $(e_i)_{i \in I}$ of $E$ and $(f_j)_{j \in J}$ of $F$ will yield a basis $(e_i \otimes f_j)_{(i,j) \in I\times J}$ of $A=E\otimes _kF$ ( This answers your question 3.)
b) The set $Spec(A)$ of prime ideals of the non-zero ring $A$ is thus non empty.
Consider a prime ideal $P \subset A$, the quotient $A/P$ and the fraction field $k(P)=Frac(A/P)$.
The compositon of $k$-morphisms $E \to A=E\otimes _k F \to A/P \to k(P) $ is necessarily injective (since $E$ is a field!) and exhibits $k(P)$ as an extension of $E$. Similarly $k(P)$ is canonically an extension of $F$. And finally $k(P)$ is a compositum of $E$ and $F$. (This answers your question 2. and part of your question 1. For the rest of 1. see c) below)
c) Every compositum of the extensions $E,F$ of $k$ is obtained (up to isomorphism) by the procedure above applied to a suitable prime $P\in Spec(A)$. Moreover different primes yield non-isomorphic extensions. So $Spec(A) =Spec(E\otimes _kF)$ exactly classifies the isomorphism classes of composita of $E$ and $F$. In particular, since $Spec(A)$ has in general more than one element, it does not make sense to talk of the compositum of $E$ and $F$ in the abstract, that is if $E$ and $F$ are not given as subfields of some extension of $k$. (This answers the rest of your question 1.)
d) A paraphrase of the above is that we have described the set underlying the affine scheme $Spec(E\otimes _k F)$. More generally, the description in c) is the key point in the description of the set underlying the fiber product $X\times _S Y$ of two arbitrary schemes $X,Y$ over an arbitrary scheme $S$.
Addendum: what is a compositum?
Given two field extensions $k\to E, k\to F$, let me explain in elementary terms (that is without tensor products) what a compositum of these is.
It is the data of a field extension $k \to K$ and of a pair of $k$-morphisms ($E\to K, F\to K$), subject to the condition that the union of the images of $E$ and $F$ in $K$ generate (in the field sense !) the field $K$. An isomorphism of two such extensions is an isomorphism $K\to K'$of $k$-extensions making obvious diagrams commute.
The key point to keep in mind is that the extension $k\to K$ alone does not determine the compositum.
This is perfectly illustrated by Matt's and Pierre-Yves's exchange in the comments where ( a little confusingly!) the three extensions $E,F,K$ of $k=\mathbb R$ are the inclusion $k=\mathbb R \to K=\mathbb C$ and they have two composita with the $same$ extension $k\to K$ which are nevertheless non-isomorphic: this is possible because the composita have different pairs of $k$-morphisms ($E\to K, F\to K$) in their data.
