Ordering the field of real rational functions

It's easier to order first $\mathbb{R}[X]$ (the ring of polynomials) and then use the fact that an ordering on a domain extends uniquely to its field of fractions.

The ordering on $\mathbb{R}[X]$ can be defined by

$$ f < g \quad\text{if and only if}\quad \lim_{x\to\infty}(g(x)-f(x))>0 $$ Where the limit can be $\infty$. This is is the same as saying that the leading coefficient of $g-f$ is positive. Checking the order properties is easy.

When $D,\le$ is an ordered domain and $F$ is its field of fractions, then there's a unique extension of $\le$ to $F$ making $F,\le$ an ordered field, by defining $$ \frac{a}{b}\le\frac{c}{d} \quad\text{if and only if}\quad ad\le bc $$ where $b,d>0$.

You’re making matters more difficult than necessary. Let $P$ be the set of rational functions with positive ratio together with the $0$ function; what you want to show is that $P$ satisfies the definition of a positive cone in the field $F$ of rational functions: it’s closed under addition and multiplication, the square of every $f\in F$ is in $P$, and $-1\notin P$. All of these are very easily verified. Now for rational functions $f$ and $g$ define $f\le_P g$ if and only if $g-f\in P$, and prove that $\langle F,\le_P\rangle$ is an ordered field according to the order-based definition (or just appeal to the easy result that the definition via a linear order is equivalent to the definition via a positive cone).

You write "Since the relation $>$ on any "pair" of real rational functions $r$, $r′$ essentially depends on the comparability of the leading-coefficient ratios...", but this is not really true. It's your choice to desire that method of ordering real rational functions, but it's not mandatory. While $\mathbf R$ has only one possible ordering, there are many orderings on $\mathbf R(x)$ besides the one you're asking about, which could be called ordering by growth at $\infty$. There is also ordering of rational functions by growth at $-\infty$ or by growth just to the right of a real number or just to the left of a real number (e.g., ordering by growth just to the right of $1$ is the rule $f > g$ when $f(1+\varepsilon) > g(1+\varepsilon)$ in $\mathbf R$ for all small positive $\varepsilon$). Since any rational function has only a finite number of poles (numbers where the denominator blows up), these orderings all make sense: for each $a \in \mathbf R$, the domain of each rational function includes $(a,a+\varepsilon)$ and $(a-\varepsilon,a)$ for small enough $\varepsilon > 0$.

The orderings I described on $\mathbf R(x)$ exhaust all possibilities, but I don't have the time to write out an explanation of that here.