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".
See exercises (6.3) and (6.4) of Cassels-Frohlich book on algebraic number theory. In these exercises, an example of two number fields $E,E'$ with the same zeta function is given; therefore, the set of primes which split in $E,E'$ are the same.
This amounts to constructing two subgroups $H,H'$ in a finite group $G$, which are not conjugate but meet every conjugacy class of $G$ in the same number of elements.
Since the OP has asked for split (but not necessarily completely split), the fact that equality of zeta functions implies that the split primes are the same needs a small argument. Let $p$ be a prime and $f_1,\cdots, f_r$ be the residue class degrees in $E$ over $p$ (in decreasing order); simlarly $g_1,\cdots, g_s$ the residue class degrees in $E'$ over $p$ (in decreasing order). Assume $f_1\geq g_1$. The local zeta functions at $p$ are
$$(1-X^{f_1})\cdots (1-X^{f_r})=(1-X^{g_1})\cdots(1-X^{g_s})$$where $X=p^{-s}$. Since the multiplicity of the root $X=1$ is the same, we get $r=s$. The cyclotomic polynomial $\Phi _{f_1}(X)$ divides the left hand side and hence the right hand side. Therefore, by the uniqueness of irreducible factors on the LHS-RHS, $g_1=f_1$. We can thus cancel the factor $1-X^{f_1}$ on both sides and use induction on $r$ to conclude that $f_i=g_i$ for all $i$.
Best Answer
If $L/K$ is a finite, Galois extension of number fields such that $\text{Gal}(L/K)$ is not cyclic, then no prime of K remains inert L. Indeed, one always has an isomorphism $D_p/I_p\cong \text{Gal}(L_p/K_p)$ of the Decomposition group modulo the Inertia group with the Galois group of the corresponding residue field extension. The latter group is the Galois group of a finite extension of finite fields, hence is cyclic. If the prime p were to remain inert in L, then by definition the Inertia group would be trivial and the Decomposition group would be all of $\text{Gal}(L/K)$. But this would imply that $L/K$ was a cyclic extension - a contradiction.
[Edit] I can't help but mention a cute application of this. Let $n$ be any positive integer for which $(\mathbb{Z}/n\mathbb{Z})^*$ is not cyclic. Then the cycotomic polynomial $\Phi_n(x)$ is reducible modulo $p$ for every rational prime $p$. Indeed, suppose that $p$ is a rational prime for which $\Phi_n(x)$ is irreducible modulo $p$. Then by the Dedekind-Kummer theorem, $p$ is inert in the cyclotomic field $\mathbb{Q}(\zeta_n)$. Then the Galois group of the residue class field extension, which is cyclic, is isomorphic to the Decomposition group, which in this case is $\text{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})\cong(\mathbb{Z}/n\mathbb{Z})^*$. But the latter group is not cyclic - contradiction. Thus $\Phi_n(x)$ is reducible modulo $p$ for all rational primes $p$.