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".
The following result holds.
Theorem.
(1) $\,$ (Baer-Kaplanski) $\,$ If $G$ and $H$ are torsion groups with isomorphic endomorphism rings $\mathrm{End}(G)$ and $\mathrm{End}(H)$, then $G$ and $H$ are isomorphic, and any ring isomorphism $\psi \colon \mathrm{End}(G) \to \mathrm{End}(H)$ is induced by some group isomorphism $\varphi \colon G \to H$.
(2) $\,$ (Leptin-Liebert) If $G$, $H$ are abelian $p$-groups $(p >3)$ and $\mathrm{Aut}(G)$ is isomorphic to $\mathrm{Aut}(H)$, then $G$ is isomorphic to $H$.
See A. V. Mikhalev, G. Pilz: The Concise Handbook of Algebra, p.74 and the references given therein.
Best Answer
If you do not impose an algebraically closed condition, no two are equivalent. This basically follows from your (3). Namely, observe that
Now as you've observed, the automorphism group of $\bar{\mathbb{F}}_p(x)$ over $\bar{\mathbb{F}}_p$ is $PGL_2(\bar{\mathbb{F}}_p)$. These groups are nonisomorphic for different primes. To see this, e.g. note that $PGL_2(\bar{\mathbb{F}}_p)$ contains a maximal abelian subgroup isomorphic to $\mathbb{F}_p$ (of unipotent upper-triangular matrices), but not $\mathbb{F}_\ell$ for $\ell \ne p$ (as all other maximal commutative subgroups are conjugate to diagonal matrices, which have lots of different prime components).
The algebraically closed case seems to be more difficult from this point of view.