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 it's possible that non-geometric extensions are indeed not as directly visualizable as geometric ones.
Some terminology: let $k$ be a field, and either assume $k$ has characteristic $0$ or beware that some separability issues are being omitted in what follows. A (one variable) function field over $k$ is a finitely generated field extension $K/k$ of transcendendence degree
one. This already allows for the possibility of a nontrivial constant extension, which is often excluded in geometric endeavors: for instance, according to this definiton, $\mathbb{C}(t)$ is a function field over $\mathbb{R}$, but a sort of weird[1] one: e.g. it has no $\mathbb{R}$-points.
One says a function field $K/k$ is regular if $k$ is algebraically closed in $K$; i.e., any element of $K$ which is algebraic over $k$ already lies in $k$ [plus separability stuff in positive characteristic]. Any function field can be made regular just by enlarging the constant field to be the algebraic closure of $k$ in $K$; e.g., the previous example is a regular function field over $\mathbb{C}$.
Regularity is what one needs to think about function fields as geometric objects: namely, there is a bijective correspondence between regular function fields $K/k$ and complete, nonsingular algebraic curves $X_{/k}$.
Now, on to covers. Let $L/K$ be a finite degree extension of function fields over $k$. One says (often; this is slightly less standard terminology) that the exension $L/K$ is geometric over $k$ if both $L$ and $K$ are regular function fields. And again, there is a bijective correspondence between geometric extensions of function fields and finite $k$-rational morphisms of algebraic curves $Y \rightarrow X$.
Assuming that the bottom function field $K$ is regular, every extension $L/K$ may be decomposed into a tower of a constant extension $lK/K$ followed by a geometric extension $L/lK$. Constant extensions have a role to play in the theory -- see for instance the chapter on constant extensions in Rosen's Number theory in function fields, but I think it is fair to describe their role as algebraic rather than geometric: at least that's the standard view.
In fact, the issue that not all extensions of regular function fields are geometric is an important technical one in the subject, because sometimes natural algebraic constructions do not preserve the class of geometric extensions.
Here is an example very close to my own heart: let $p$ be an odd prime. The elliptic modular curves $X(1)$ and $X_0(p)$ have canonical models over $\mathbb{Q}$ and there is a natural "forgetful modular" covering $X_0(p) \rightarrow X(1)$. This corresponds to a geometric extension of function fields $\mathbb{Q}(X_0(p)) / \mathbb{Q}(X(1))$. This is not a Galois extension: what is the Galois closure and what is its Galois group? If -- as was classically the case -- our constant field were $\mathbb{C}$ -- then the Galois closure is the function field of the modular curve $X(p)$ and the Galois group of the covering $X(p)/X(1)$ is
$\operatorname{PSL}_2(\mathbb{Z}/p\mathbb{Z})$. However, over $\mathbb{Q}$ the Galois closure also contains the quadratic field $\mathbb{Q}\left(\sqrt{(-1)^{\frac{p-1}{2}} p}\right)$ so is an extension of a cyclic group of order $2$ by $\operatorname{PSL}_2(\mathbb{Z}/p\mathbb{Z})$ (in fact it is $\operatorname{PGL}_2(\mathbb{Z}/p\mathbb{Z})$). Thus the extension is not geometric. This is unfortunate, because Hilbert's Irreducibility Theorem says that if one has a geometric Galois extension $L/k(t)$ with $k$ a number field, then one can realize $\operatorname{Aut}(L/k(t))$ as a Galois group over $k$. So in this case, this obtains $\operatorname{PSL}_2(\mathbb{Z}/p\mathbb{Z})$ as a Galois group over not $\mathbb{Q}$ but over the variable quadratic field given above. K.-y. Shih found a brilliant way to "tweak" this construction to realize $\operatorname{PSL}_2(\mathbb{Z}/p\mathbb{Z})$ over $\mathbb{Q}$ in certain (infinitely many) cases, and other mathematicians -- e.g. Serre, myself, my graduate student Jim Stankewicz -- have put a lot of thought into extending Shih's work, but with only very limited success.
Added: Brian's example in the comments is very nice. Maybe another remark to make is that in the arithmetic theory of coverings of curves (an active branch of arithmetic geometry) the distinction between a Galois extension and a geometrically Galois extension of fields (i.e., one which becomes Galois after base change to $\overline{k}$) is a key one: it's certainly something that many arithmetic geometer think a lot about. It just doesn't come with an obvious "visualization", at least not to me. Not everything in algebraic or arithmetic geometry can be visualized, or at least not visualized in a way common to different workers in the field. For instance, an inseparable field extension $l/k$ is by definition ramified, but I have never seen anyone describe this visually. (There are things you can say to justify that this is not a "covering map", e.g. by pointing to the nonreducedness of $l \otimes_k l$, but I don't think this is direct visualization either. Maybe some would disagree?) What you do is think of the case of a ramified cover of Riemann surfaces, and take away the (key) piece of intuition that an inseparable field extension -- which is, visually speaking, just one closed point mapping to another -- behaves like a ramified cover of Riemann surfaces in many ways. So, as Brian says, in this subject a lot of geometric reasoning proceeds by analogy. Unlike in, say, certain branches of low-dimensional topology, one does not prove a theorem by referring to (allegedly) visually apparent features of one's constructions.
[1]: Those who know me well know that I certainly don't think that a curve is weird just because it has no degree one closed points. More accurate is to say that this curve doesn't have any degree one closed points for a "weird reason".
Best Answer
I think you are lumping too many disparate kinds of fields together under the heading "zero-dimensional". As Jason says in his answer, there are some precise definitions of dimensions of fields (e.g. cohomological dimension but also other definitions of a field-arithmetic nature).
Another important comment is that in modern algebraic / arithmetic geometry there is no restriction on the kind of field that can be taken as a "ground field", i.e., over which to define algebraic varieties. Although you have indeed listed some common ground fields, some people think about other special cases as well as the general case. (For instance I have recently been thinking about elliptic curves defined over transfinitely iterated function fields.)
The basic classification of fields, as I understand/view it, is as follows:
Step 1: Every field has a characteristic, either $0$ or a prime number. There are no homomorphisms between fields of different characteristics, so fields of a given characteristic are somehow different worlds (one can think of the characteristic as a connected component in the category of fields, essentially).
Step 2: For a fixed characteristic $p \geq 0$ there is a unique minimal field, called the prime subfield, say $k_0$. This is precisely the initial object in the category of fields of characteristic $p$: it's $\mathbb{Q}$ in characteristic $0$ and $\mathbb{F}_p$ in characteristic $p$. This means that the absolute theory of fields is reduced to the theory of field extensions, and one can define "absolute invariants" on a field $K$ by giving invariants of $K/k_0$. In particular:
Step 3: The absolute transcendence degree of a field $K$ is the cardinality of a transcendence basis for $K/k_0$. This means that there is a subextension $k_0 \subset F \subset K$ such that $F/k_0$ is purely transcendental: a rational function field, perhaps in infinitely many variables, and $K/F$ is algebraic.
Step 4: The algebraic extension $K/F$ is in many ways the most interesting part. For instance, as Qiaochu Yuan mentions, algebraic geometry in dimension $n$ is the study of finite degree field extensions of $\mathbb{C}(t_1,\ldots,t_n)$.
When $n = 1$ we get precisely the compact Riemann surfaces, or (replacing $\mathbb{C}$ with any field $k$ and looking at finitely generated field extensions of transcendence degree $1$) of complete, regular, integral algebraic curves over $k$. So a classification here is the (rich) classical story of the genus, the moduli spaces $\mathcal{M}_g$, and so forth. When $k$ is not algebraically closed, Galois cohomology enters the picture.
When $n \geq 2$ we are studying birational algebraic geometry only, but that is still very rich. When $n = 2$ there is a "classification of algebraic surfaces" which ends up tossing most of them into a very large box called "general type". To the best of my knowledge there is no explicit description of the connected components of the (infinite type) moduli space of all complex algebraic surfaces like there is in dimension one.
In fact, I believe that probably starting even in dimension $2$ it is in some sense hopeless to try to give an algorithmic classification, although I cannot recall having seen a precise impossibility theorem along these lines analogous e.g. to the algorithmic impossibility of classifying compact $4$-manifolds because this problem -- via the fundamental group -- contains the word-problem for finitely presented groups which Novikov proved is algorithmically unsolvable. Bjorn Poonen discussed similar (open) problems in the last of a series of three lectures on undecidability that he gave at UGA a few years ago; notes are available on his webpage.
Step 5: A vaguely dual role to purely transcendental extensions of the prime subfield is played by the algebraically closed fields. Here there is a precise classification: the characteristic and the absolute transcendence degree determine an algebraically closed field up to isomorphism. If the field is uncountable, then the absolute transcendence degree is just its cardinality, so that classifies the field: e.g. the only algebraically closed field of characteristic $0$ and continuum cardinality is $\mathbb{C}$. This has been used for some sneaky purposes: e.g. the algebraic closure of $\mathbb{Q}_p$ must then also be isomorphic to $\mathbb{C}$. (However for countable fields cardinality is not enough: e.g. $\overline{\mathbb{Q}}$ is not isomorphic to $\overline{\mathbb{Q}(t)}$.) This uncountable categoricity means that the first order theory of algebraically closed fields of given characteristic $p \geq 0$ is complete, which has various pleasant consequences. Algebraically closed fields of infinite transcendence degree have some further nice properties which makes them suitable for use as "ground fields" in algebraic geometry, as was exploited by Weil in his pre-(scheme-theoretic) foundations. From a model-theoretic perspective, these large algebraically closed fields enjoy good saturation properties.
Fields which are "close" to being algebraically closed tend to be better understood than fields which are farther away from being algebraically closed. One can measure this (at least in characteristic $0$) by the size / complexity of the absolute Galois group $\operatorname{Aut}(\overline{K}/K)$. Thus for instance the absolute Galois group of $\mathbb{Q}_p$ is known completely -- i.e., it is a topologically finitely generated compact totally disconnected Hausdorff group, and explicit generators and relations are known, which in some sense means we understand every algebraic extension of $\mathbb{Q}_p$. This needs to be taken with a grain of salt: a certain filtration on the absolute Galois group provides much coarser information -- e.g. you can break the group up into three pieces, two of which are commutative and one of which is pro-$p$, so the group is pro-solvable. That's useful: the pro-$p$-part is the hard part and knowing generators and relations for it does not in practice seem to take away the mystery. For instance, the Local Langlands Correspondence is a deep theorem about representations of the absolute Galois group of a $p$-adic field, and it was certainly not proved by looking at its explicit structure as a topologically finitely presented group!
Let me note finally that every compact totally disconnected topological group occurs up to isomorphism as the automorphism group of an algebraic Galois extension of fields (the Leptin-Waterhouse Theorem; I discovered this independently as a graduate student, and at the time I knew neither that others had published the result before nor even that such a result would be publishable)...so algebraic field extensions can be awfully complicated.