The problem with your “foo extension” is that the group structure of $H$ does not matter! Even in the finite case, you are treating the “group” $H=\mathbb{Z}_k$ (with $k$ the index of $F$ in $G$) as simply a set of indices. As such, all you are doing is picking coset representatives. Note that your construction stops working if the index has cardinality greater than $\aleph_0$, since a cyclic group must necessarily be countable.
So... your notion of “foo extension” is really just that of subgroup/overgroup. You have that $G$ is a foo extension of $F$ if and only if $F$ is a subgroup of $G$, if and only if $G$ is an overgroup of $F$. Indeed, if you are going to admit the axiom of choice, recall that under the Axiom of Choice, every nonempty set has a group structure (and in fact, this is equivalent to the Axiom of Choice). Thus, given any subgroup $F$ of a group $G$, let $X$ be a set of left coset representatives of $F$ in $G$, choosing $e_g$ for the coset $F$. Give $X$ an arbitrary group structure that makes $e_G$ the identity, $H=(X,\cdot)$ (to see this is possible, give $X$ an arbitrary group structure, and then permute $e_G$ with the identityh of the group structure and use transport of structure). Now define your map $\psi(F,H)\to G$ as $\psi(f,x) = fx$, as you do.
Conversely, the existence of $\psi$ and the requirement that $\phi=\psi(\cdot,e_H)$ be a homomorphism (injectivity follows from the requirement that $\psi$ be a bijection) yields that $F$ is (isomorphic to) a subgroup.
Thus, your concept of “foo extension” is identical to that of subgroup/overgroup.
Now, your “symmetric foo extension” is exactly the same as usual notion strict factorization of a group: $G=FH$ with $F\cap H=\{e\}$. Indeed, if you can express $G$ as a product of two subgroups, $G=FH$ with $F\cap H=\{e\}$, then the bijection $\psi\colon F\times H\to G$ given by $\psi(f,h) = fh$ is the desired function; the restrictions of $\psi$ to both $F\times\{e\}$ and $\{e\}\times H$ are homomorphisms, since $\psi(f_1f_2,e) = f_1f_2 = (f_1e)(f_2e) = \psi(f_1,e)\psi(f_2,e)$, and similarly for $\psi(e,h_1h_2)$. Conversely, the existence of your bijection and the fact that the restriction of $\psi$ to $F\times\{e\}$ and to $\{e\}\times H$ are homomorphisms shows that $G=FH$ with $F$ and $H$ (isomorphic to) subgroups of $G$; and the fact that each element of $G$ is expressible uniquely this way implies that $F\cap H=\{e\}$.
However, you are not ensured normality of either subgroup. For an example of that, take $G=A_5$ the alternating group of $5$ elements, which has order $60$. Let $F$ be a subgroup isomorphic to $A_4$ of order $12$, for example the subgroup of $A_5$ that fixes $5$. Let $H$ be a subgroup of order $5$, for example, the one generated by the $5$-cycle $(1,2,3,4,5)$. Then $|FH|=|F||H|/|F\cap H| = 60$, hence $FH=A_5$; however, neither $F$ nor $H$ are normal in $A_5$ (as $A_5$ is simple).
Added. From the point of view of the subgroup $F$ (or $H$) the symmetric foo extension is the notion of complement: given a group $G$, a subgroup $F$ has a complement if there exists a subgroup $H$ such that $G=FH$ and $F\cap H = \{e\}$. Note however, that complements need not be unique (above, any $5$-cycle will give a complement of $H$ in $A_5$), and they need not be isomorphic. They are related to the notion of Zappa-Szep product.
So you do not obtain the usual notion of of extension of $F$ by $H$ (careful with the nomenclature! Sometimes it is expressed the other way; see the discussion here )
Answer 0. I do not think you will be able to express it in the form you desire, unless you put conditions on $\psi$ that artificially encode the normality of $F$. As to why they are useful, yes, they are useful for decomposition, but more generally, normal subgroups are intimately connected with homomorphisms, and with the notion of congruence, and fit into a much more general framework of Universal Algebra. See this previous question.
Answer 1. The definition of extension provides slightly more information than just saying that $F\triangleleft G$. When you say that $G$ is an extension of $F$ by $H$, you are saying that $F\triangleleft G$, and that $G/F$ is isomorphic to $H$. That is, you are also describing the isomorphism type of the quotient.
The reason for a “separate definition” is that sometimes you are interested in the normal subgroups alone (when studying the structure of the group, when considering congruences, etc) but sometimes you are also interested in the structure of the quotient (for example, when doing cohomology or representation theory). The definition of “normal” only tells you a little about how $F$ sits inside $G$, the definition of “extension of $F$ by $H$” tells you that plus the structure of $G/F$.
More importantly, one usually approaches the notion of “extension of $F$ by $H$” from the other direction: you know who $F$ and $H$ are, and the only thing you know about $G$ is that it has a normal subgroup isomorphic to $F$ and the quotient is isomorphic to $H$. That is, you are trying to understand $G$ in terms of $F$ and $H$; in the definition of normal subgroup, you usually already know $G$.
Answer 2. Since your “symmetric foo extension” is equivalent to saying that $G$ can be factored as $G=FH$ with $F\cap H=\{e\}$, if you also add the requirement that $F\triangleleft G$, then yes, you get that a split extensions is equivalent to a “symmetric foo extension with the requirement that the image of $\psi(\cdot,e) = \phi$ be normal in $G$.
Split extensions are the “trivial” case of the extension (they give semidirect products). When studying extensions in general, they are the “easy” case.
Perhaps the reason for the interest in extensions and split extensions can be made clear with a bit of history.
Schreier proposed a programme for studying finite group. One key to the programme is what is known as the Jordan-Holder Theorem. It says that every finite group can be “decomposed” in the following sense: you can find a sequence of subgroups
$$1=N_0\lt N_1\lt N_2\lt N_3\lt\cdots\lt N_k=G$$
such that $N_i\triangleleft N_{i+1}$ and with $N_{i+1}/N_i$ simple (has no normal subgroups other than the trivial and the whole subgroups). If you then take the multiset of simple groups $N_{1}/N_0,\ldots,N_k/N_{k-1}$, this list is unique up to order for any such decomposition. Thus, there is a list, with multiplicities, of “subfactors” of $G$ that are uniquely determined by $G$; kind of like prime factorization of integers.
So Schreier proposed the following programme:
- Describe all finite simple groups.
- Assuming we understand $F$ and $H$, describe all possible extensions of $F$ by $H$.
In principle, 2 is a finite problem.
Now, there are some issues with this. The description of all finite simple groups turns out to be pretty complex (over ten thousand pages in hundreds of papers by dozens of authors, first announced in the early 80s, then found wanting and corrected in the mid-90s, still being published in single-book form, currently in its “second generation”, with an effort of a “third generation proof” using fusion systems). And the description of all extensions of $F$ by $H$ also turns out to be pretty complicated. The easy case is that when the extension is split. All other cases are coded via the second cohomology group, which is difficult to calculate even in the “simpler” case where $F$ is abelian. But because of this idea that if you can describe both $F$ and $G/F$, and how $F$ and $G/F$ are put together, you can get some information about $G$, you have an interest in such constructions.
The second cohomology group encodes the obstructions to the extension being split; the trivial element corresponds to the split extensions.
