Let me address your first question. First, I want to argue that there is no precise meaning of "involving numbers only". For example, given a finite field $F$ of size $4$ constructed in the usual manner (quotient of a polynomial ring over $\mathbb{Z}/2\mathbb{Z}$), I can choose a set of numbers, say
$$S=\{37,\tfrac{5}{19},\pi,e\}$$
and, choosing a bijection of $S$ with $F$, use transport of structure to give $S$ the structure of a field. The field structure does not depend in any way on what the underlying set is "made of".
However, along the lines of what I think you are ultimately after, you can obtain finite fields of any possible order using larger rings of integers. For example, $\mathbb{Z}[i]/(3)$ is a finite field of size $9$, and $\mathbb{Z}[i]$ consists of very reasonable numbers,
$$\mathbb{Z}[i]=\{a+bi\mid a,b\in\mathbb{Z}\}.$$
Now let me addressr your second question. Let's use $\mathbb{F}_p$ to mean $\mathbb{Z}/p\mathbb{Z}$, a finite field of order $p$ - it is a very common notation that is slightly less cumbersome, but doesn't mean anything different, they are exact synonyms.
A finite field of order $p^n$ is often constructed by taking the polynomial ring $\mathbb{F}_p[x]$, choosing an irreducible polynomial $f\in \mathbb{F}_p[x]$ of degree $n$, and then making the field
$$F=\mathbb{F}_p[x]/(f).$$
Now, the division algorithm for polynomials tells you that each equivalence class in this quotient can be uniquely identified by a representative of degree $<n$. In other words,
$$\begin{align*}
F&=\{a_0+a_1x+\cdots +a_{n-1}x^{n-1}+(f)\mid a_0,a_1,\ldots,a_{n-1}\in\mathbb{F}_p\}\\\\
&=\left\{\,\overline{a_0+a_1x+\cdots +a_{n-1}x^{n-1}}\,\;\middle\vert\;a_0,a_1,\ldots,a_{n-1}\in\mathbb{F}_p\right\}\\\\\\
&=\{a_0+a_1\overline{x}+\cdots +a_{n-1}\overline{x}^{n-1}\mid a_0,a_1,\ldots,a_{n-1}\in\mathbb{F}_p\}
\end{align*}$$
Letting the symbol $\alpha$ be a stand-in for $\overline{x}$, you can think of $F$ as being $\mathbb{F}_p$ with a new element "$\alpha$" added in, where $\alpha$ is a root of $f$, and you can write $F=\mathbb{F}_p[\alpha]$.
Now, the prime subfield of $F$ is just the "constant" polynomials, i.e. the ones with no $\alpha$'s in them:
$$\text{the prime subfield of }F=\{a_0+0\alpha+\cdots+0\alpha^{n-1}\mid a_0\in\mathbb{F}_p\}$$
and for each divisor $d\mid n$, the unique subfield of $F$ of order $p^d$ is the collection of polynomials in $\alpha$ whose terms are those of exponents that are multiples of $n/d$:
$$\text{the subfield of }F\text{ of order }p^d=\{a_0+a_1\alpha^{n/d}+\cdots+a_{d-1}\alpha^{(d-1)n/d}\mid a_0,a_1,\ldots,a_{d-1}\in\mathbb{F}_p\}$$
(clearly, the above set has cardinality $p^d$, because it takes $d$ elements of $\mathbb{F}_p$ to specify a given element of the above set, namely, each of the coefficients of the powers of $\alpha$. To see that it is a field, remember that $(a+b)^p=a^p+b^p$ in a field of characteristic $p$.)
Any ordered field $F$ has characteristic $0$, so it contains a copy of $\mathbb{Z}$; by the universal property of the quotient field, the ring monomorphism $\mathbb{Z}\to F$ lifts to a monomorphism $\mathbb{Q}\to F$. We can identify $\mathbb{Q}$ with its image, so it's not restrictive to assume that $\mathbb{Q}\subseteq F$.
It's not really difficult: if $m/n\in\mathbb{Q}$, then we send it to
$$
\frac{f(m)}{f(n)}\in F
$$
where $f\colon \mathbb{Z}\to F$ is the (unique) monomorphism. Is this a field homomorphism? Just a check.
Now we come to the order. First of all, positive integers are positive in $(F,\prec)$: if $n>0$, then
$$
n=\underbrace{1+1+\dots+1}_{\text{$n$ times}}
$$
and therefore $0\prec n$. Conversely, if $n<0$, then
$$
n=-(\,\underbrace{\,1+1\dots+1}_{\text{$-n$ times}}\,)
$$
and so $n\prec0$.
Any element of $\mathbb{Q}$ can be represented as $m/n$ with $n>0$, because $a/b=(-a)/(-b)$, where $a,b\in F$, $b\ne0$. So, let $0\prec m/n$ in the ordering of $F$, with $n>0$. Then, by the properties of ordered fields,
$$
0\prec n\cdot\frac{m}{n}=m
$$
and therefore $m>0$. So a rational which is positive in $(F,\prec)$ is also positive in the usual order. A rational which is negative in $(F,\prec)$ is the opposite of a positive rational (in both orders).
Best Answer
Let $K$ be a subfield of $\Bbb R$ which is isomorphic to $\Bbb R$. Then $K$ has a canonical ordering: $x\ge y$ iff $x=y+t^2$ for some $t\in K$. As $K$ is order-isomorphic to $\Bbb R$, then $K$ is a complete ordered field, where we take the completeness axiom as "a nonempty bounded-above subset has a supremum".
The embedding of $K$ in $\Bbb R$ must be order preserving. Suppose $\xi\in\Bbb R$. The set $S=\{a\in \Bbb Q: a<\xi\}$ has a supremum in $K$ and in $\Bbb R$, but these must be the same, namely $\xi$. So $\xi\in K$ for all $\xi\in\Bbb R$.