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).
There are two ways you might want to represent a field of order $q^k$ (where $q\in \mathbb{N}$ is a prime and $k \gt 0$ a positive integer).
One is to imagine a simple algebraic field extension of $\mathbb{Z}_q$ which has a basis $\{1,\alpha,\ldots,\alpha^{k-1}\}$, namely $\mathbb{Z}_q[\alpha]$ where $\alpha$ satisfies an irreducible monic polynomial of degree $k$ over $\mathbb{Z}_q$:
$$ \alpha^k + c_1 \alpha^{k-1} + \ldots + c_{k-1}\alpha + c_k = 0 $$
Any element of this field extension can be represented in terms of the basis, and the addition is just the same as the addition of the $k$-dimensional vector space over $\mathbb{Z}_q$. But when we multiply, we have to perform substitutions to eliminate powers $\alpha^k$ and higher by using:
$$ \alpha^k = - c_1 \alpha^{k-1} - \ldots - c_{k-1}\alpha - c_k $$
The other way to think about it is to start with an irreducible polynomial $p(x)$ of degree $k$ over $\mathbb{Z}_q$ and construct $\mathbb{Z}_q[x]/p(x)$ as a quotient ring. Since the ideal generated by $p(x)$ is maximal, the quotient ring is a field and has dimension $k$ over $\mathbb{Z}_q$ as a vector space.
These two constructions are equivalent, with $\alpha$ being a root of:
$$ p(x) = x^k + c_1 x^{k-1} + \ldots + c_{k-1}x + c_k $$
and identified with $x \bmod{p(x)}$ in $\mathbb{Z}_q[x]/p(x)$.
It turns out that all field extensions of degree $k$ over $\mathbb{Z}_q$ are isomorphic, so as a computational convenience the irreducible polynomial $p(x)$ may be chosen to be simple in some way (e.g. having as few nonzero coefficients as possible).
It's not hard to come up with a monic irreducible polynomial of degree $3$ over $\mathbb{Z}_2$ in order to construct a field of $8$ elements. You only need to check for divisibility by linear (first degree) factors to be sure of irreducibility.
Best Answer
Whilst the product of two fields is never a field, an ultraproduct always is. If you try taking an ultraproduct of two fields, or any finite number of fields, then you will just get one of the fields back.
However, things get exciting when you take an ultraproduct of infinitely many fields. For example if you take an ultraproduct of all fields of prime order: $\mathbb{F}_2, \mathbb{F}_3,\mathbb{F}_5,\cdots$, you will get a field of characteristic $0$. This goes against the intuition mentioned in comments that fields of different characteristics do not talk to each other.
In simple terms, if we have an infinite sequence of fields $k_1,k_2,k_3,\cdots$, we can first take their product, and then say that two elements of this product are the same if and only if they agree on a large set of components. The result is a field.
What do we mean by large though? Given a subset of $\mathbb{N}$ we need to define what it means to be large. Then given two elements in the product, we can look at the set of indices where they agree, and we will know if they are the same element in our field based on whether or not this set is large.
Unfortunately, we cannot actually specify what we mean by large as we need the axiom of choice to prove that such a concept exists. What we can do is specify what properties we want largeness to satisfy, and then prove using the axiom of choice that such a notion exists (in fact many such notions).
Properties:
Being large cannot be equivalent to containing some number $i$. If it was then we would just get $k_i$ as the result of our ultraproduct.
If $A\subseteq B$ and $A$ is large, then $B$ is large.
If $A,B$ both large, then $A\cap B$ large.
If $A$ not large, then its complement $A^c$ is large.
The empty set $\emptyset$ is not large.
Note that if we partition a large set $A$ into two pieces $B,C$, one of them must be large, as otherwise by (2) and (3) we would have $A\cap B^c\cap C^c=\emptyset$ large, contradicting (4).
If a single element set was large, then any set containing that element would be large by (1), and any set not containing the element would not be large (by (2) and (4)), so we would contradict (0).
If a finite set were large, we could keep removing one element at a time, till we reached the empty set. At each stage we have a large set, as either we have a large set, or the singleton set we removed is large, which cannot happen. This contradicts (4).
Proof that we get a field: Let $K$ denote the product of the $k_i$ and let $I$ denote the set of elements which are $0$ on a large set of components. We know $I$ is closed under addition by (2). We know it is closed under multiplication by elements of $K$ by (1). Thus $I$ is an ideal, and our ultraproduct is just the ring $K/I$. Note this ring is non-zero by (4).
To see that this ring is a field, take any element $x\in K$. Let $y\in K$ be the inverse of $x$ on every component where $x$ is non-zero, and $0$ elsewhere. Then by (3), either $x$ is $0$ on a large set, or $xy$ agrees with the identity $(1,1,1\cdots)$ on a large set. That is every non-zero element of $K/I$ has an inverse.
Finally lets return to our example: an ultraproduct of all fields of prime order, $\mathbb{F}_2, \mathbb{F}_3,\mathbb{F}_5,\cdots$. To see that it has characteristic $0$, just note that $(m,m,m,\cdots)$ agrees with $0$ only on the finite set of components $\mathbb{F}_p$ where $p|m$. Finite sets cannot be large, so we have $(m,m,m,\cdots)\neq 0$ in $K/I$.