I'm taking a first course in Model Theory. There's a kind of argument that is repeatedly used by my lecturer but I just can't get my head around it.
Example
Hilbert's 17th problem: Every positive semidefinite polynomial over $\mathbb{R}$ is a sum of squares of rational functions.
I paraphrase a proof given in David Marker's "Model Theory: An Introduction".
We prove the contrapositive. Let $F$ be the field of rational functions over $\mathbb{R}$, with variables $x_1,\ldots, x_n$ say, and $f\in \mathbb{R}[x_1,\ldots,x_n]$ a real polynomial that cannot be expressed as a sum of squares in $F$. It follows that there is an ordering on $F$ that makes $F$ into an ordered field and $f<0$ (this is by Zorn's Lemma). Embed $F$ into its (essentially) unique real closure $R$ that preserves the order of $F$. I'm OK with everything so far.
As $f<0$ in $R$ we have $R \models \exists \bar{v}. f(\bar{v}) < 0$ (I'm not OK with this). But all real closed fields are elementarily equivalent (this is OK), so $\mathbb{R} \models \exists \bar{v}. f(\bar{v}) < 0$ and we are done.
Problem
We work in the language of ordered fields with signature $(0,1,+, \times,<)$. We have that $f(\bar{x})$ is an element of $R$, and it seems to me that the proof relies on confusing the "field variables" $\bar{x}$ with some "model-theoretic variables" $\bar{v}$. In the language of ordered fields $f(\bar{v})$, where $\bar{v}$ are model-theoretic variables, is not generally a term: indeed, terms in this language are just polynomials with integer coefficients. It follows that $\exists \bar{v}. f(\bar{v}) < 0$ is not a sentence in the language, so saying that $R \models \exists \bar{v}. f(\bar{v}) < 0$ is nonsense. My first instinct was to replace the coefficients of $f$ by more model-theoretic variables and add a few quantifiers but I still can't get the above proof to work.
Question
I've seen this strategy of pretending a polynomial is a term a few times (in model-theoretic proofs of Hilbert's Nullstellensatz and Chevalley's Theorem, for example) but I can't see what is going on. My question(s) is then
How to rigorously justify this sort of argument? Have I correctly understood Marker's proof?
I'm assuming in your question you meant $f \in \mathbb{R}[x_1, \ldots, x_n]$.
You are correct that $f$ is not necessarily given by a term in the language of ordered fields. There is, however, a term $t(\overline{x}, \overline{y})$, where $\overline{y}$ is a tuple of variables, along with some elements $\overline{a}$ from $\mathbb{R}$, such that $t^{\mathbb{R}}(\overline{b}, \overline{a}) = f(\overline{b})$ for every $\overline{b}$ from $\mathbb{R}$. The point here is that the elements $\overline{a}$ are the coefficients of $f$. We then (somewhat sloppily) write $\mathbb{R} \models \exists \overline{v}~f(\overline{v}) < 0$ to mean $\mathbb{R} \models \exists \overline{v}~t(\overline{v}, \overline{a}) < 0$. For example, if $f(x) = x^2+\pi x + \sqrt{2}$, then we have the term $t(x, y_1, y_2) := x\cdot x+y_1\cdot x+y_2$, and we see that $t^\mathbb{R}(b, \pi, \sqrt{2}) = f(b)$ for every $b \in \mathbb{R}$.
We have $\mathbb{R} \subseteq R$, so we can still make sense of $t(\overline{x}, \overline{a})$ in $R$, and in particular we can write $R \models \exists \overline{v}~f(\overline{v}) < 0$ as a (again, somewhat sloppy) way of writing $R \models \exists \overline{v}~t(\overline{v}, \overline{a}) < 0$.
Now there is an error in your paraphrase of Marker's argument. He does not use the completeness of RCF at this point, he in fact uses the model completeness, and more specifically the fact that $\mathbb{R} \preceq R$ (in the language of ordered fields). This is much stronger than saying $\mathbb{R} \equiv R$, because it tells us that truth is preserved even for statements involving parameters from $\mathbb{R}$. In particular, $R \models \exists \overline{v}~t(\overline{v}, \overline{a}) < 0$, and the parameters $\overline{a}$ come from $\mathbb{R}$, so we have $\mathbb{R} \models \exists \overline{v}~t(\overline{v}, \overline{a}) < 0$ as well.
If you prefer, another way to interpret all of this is that $\exists \overline{v}~f(\overline{v}) < 0$ is a sentence in the language with new constant symbols added for the coefficients of $f$. The fact that $\mathbb{R} \preceq R$ in the language of ordered fields means that $\mathbb{R}$ is elementarily equivalent to $R$ not just as ordered fields, but also in the language with a new constant for every element of $\mathbb{R}$, so in particular again we have that the truth of sentences mentioning $\overline{a}$ is preserved.