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 $p$.)
The function you defined is not a homomorphism. For example, $\bar{\phi}(\frac{1}{2}+\frac{1}{3})$ will be $5$ by your definition, which is clearly not $1+1$. (doesn't matter what $K$ is)
Note that $\mathbb{Q}$ is the field of fractions of $\mathbb{Z}$. We have the inclusion $i:\mathbb{Z}\to\mathbb{Q}$, $i(n)=\frac{n}{1}$. So now given an injective ring homomorphism $\phi:\mathbb{Z}\to K$ it is very natural to define:
$\bar{\phi}(\frac{m}{n})=\phi(m)\cdot[\phi(n)]^{-1}$ where $m,n\in\mathbb{Z}, n\ne 0$
Note that since $\phi$ is injective we have $\phi(n)\ne 0$, and so $\phi(n)^{-1}$ is indeed defined. Now let's check that this $\bar{\phi}$ is well defined. Suppose $\frac{m}{n}=\frac{k}{l}$. Then $ml=kn$ and so $\phi(m)\phi(l)=\phi(k)\phi(n)$. Thus:
$\bar{\phi}(\frac{m}{n})=\phi(m)\cdot [\phi(n)]^{-1}=\phi(k)\cdot [\phi(l)]^{-1}=\bar{\phi}(\frac{k}{l})$
So $\bar{\phi}$ is indeed a well defined function. Now similarly you can check that it is indeed a homomorphism and it extends $\phi$. It is also trivially injective, because it is a homomorphism of fields.
We indeed get a universal property of the field of fractions here. Suppose $R$ is an integral domain, $F$ is its field of fractions. Let $i:R\to F$ be the inclusion $i(r)=\frac{r}{1}$. Given an injective homomorphism $\phi:R\to K$ where $K$ is a field, there is a unique homomorphism $\bar{\phi}:F\to K$ such that $\bar{\phi}\circ i=\phi$. This universal property is a special case of the universal property of ring localizations.
Best Answer
Let's call the subfield from the "weird" definition $F_C$, the field generated by $1_K$ as $F_I$, and the intersection of all subfields as $F_S$. Proof that these are all equal:
Since $1_k\in F_C$, we have $F_I\subseteq F_C$, and $F_S\subseteq F_I$ by definition.
If we ask for the minimum field containing $1_K$, then we need to include $1_K+1_K$, and $1_K+1_K+1_K$, etc. For all such elements, we also need their additive inverse to be in $F_I$. That means we can define $f':\mathbb{Z}\rightarrow F_I$ in the same way as you defined $f:\mathbb{Z}\rightarrow K$, and this is a ring homomorphism. In fact, $f'(n)=f(n)$ for all $n$. So $f'$ is injective if and only if $f$ is, meaning we can lift $\overline{f}'$ into $F_I$ into the same way. But since the action of $f'$ is the same as $f$, then $\overline{f}'(x)=\overline{f}(x)$ for all $x\in\mathbb{Q}$ (or $\mathbb{Z}/p\mathbb{Z}$), since the lifting $\overline{f}$ is unique. This means:
$$ F_C=\overline{f}(\mathbb{Q}) = \overline{f}'(\mathbb{Q})\subseteq F_I$$
(or replace $\mathbb{Q}$ with $\mathbb{Z}/p\mathbb{Z}$).
Finally, note that any subfield $K'$ of the field $K$ must contain $1_K$, so it must contain the field that $1_K$ generates. Thus, $F_I\subseteq K'$ for all subfields. This means $F_I\subseteq F_S$, the intersection of all subfields.
Intuitively, $1_K$ is going to generate an additive group, and this is going to look like $\mathbb{Z}$ or $\mathbb{Z}/p\mathbb{Z}$. In the second case you're done because it will also be multiplicatively closed. In the first case you will also need the inverses of all those elements, which will look like fractions in $\mathbb{Q}$. Specifically, the field of fractions (the minimal field containing such a ring) will be isomorphic to $\mathbb{Q}$, the field of fractions of $\mathbb{Z}$.