This has a lot to do with language. $V(X)$ is an $\mathcal{L}$-structure of language $\mathcal{L}=\{=,\in,V(X)\}$ where $"="$ and $"\in"$ are predicates and $V(X)$ is a set of constants.
For $V(X)$ to be an $\mathcal{L}$-structure you need an interpretation of your language. So for every constant $x\in V(X)$ we need to associate an element $y\in V(X)$, clearly we can pick $y=x$. For our predicates $"\in"$ and $"="$ we have to a find subsets $M_{\in},M_{=}\subset V(X)\times V(X)$. We then say for $"a\in b"$ or $"c=d"$ are true if and only if $(a,b)\in M_{\in}$ and $(c,d)\in M_{=}$ respectively. Again you can pick the obvious subsets $M_{=}=\{(x,x)\in V(X)\times V(X)$ and $M_{\in}=\{(x,y)\in V(X)\times V(X):x\in y\}$. These symbols along with logical symbols as $\rightarrow,\wedge,\neg,\vee,\forall$, etc are the only symbols we may use to write statements.
The statement $(\forall x\in X)(\forall z\in x)(z\in y)$ where $y$ is some constant would be rewritten as
$$(\forall x)(x\in X\rightarrow (\forall z)(z\in x\rightarrow z\in y))$$
where $y$ and $X$ are constants.
Now the question remains, is this formula true. This does depend on your interpretation of the $"\in"$ predicate. But if you interpret it in such a way that $(x,y)\not\in M_{\in}$ for all $y\in X$, then this is a true statement. $z\in x$ is not a true for $x\in x$, so $(\forall z)(z\in x\rightarrow z\in y)$ is true by definition of $\rightarrow$ This then makes $(\forall x)(x\in X\rightarrow (\forall z)(z\in x\rightarrow z\in y))$ true again due to the definition of $\rightarrow$.
Basically this statement is true in the same way that the statement: "every element of the empty set is a vampire" is true. You will never be able to avoid these kind of silly true formula's, they are just a consequence of writing a rigorous logical language, but they are true in both $V(X)$ and set theory.
1.
Your "extensibility" condition is strictly stronger than being a real-closed field.
You already know that every structure which satisfies your extensibility condition is a real-closed field. I will now show that the converse does not hold.
Consider e.g. the theory of the structure $\mathfrak{R} = (\mathbb{R}, +, -,\cdot, \sin, 0, 1)$. Clearly $\mathfrak{R} \models \exists x. x \neq 0 \wedge \sin(x) = 0$. Moreover, by the transcendence of $\pi$ we have that for any polynomial $p$ with integer coefficients, $\mathfrak{R} \models \forall x. x \neq 0 \wedge \sin(x) = 0 \rightarrow p(x) \neq 0$. Since $p$ has integer coefficients, this is a first-order sentence in the language of fields.
But this means that the field $\overline{\mathbb{Q}} \cap \mathbb{R}$ of real algebraic numbers does not satisfy the extension property. If it did, $\sin: \mathbb{R} \rightarrow \mathbb{R}$ would have to "extend" to a function $~^\star\!\sin: \overline{\mathbb{Q}} \cap \mathbb{R} \rightarrow \overline{\mathbb{Q}} \cap \mathbb{R}$ that takes on a transcendental value. That's of course not possible.
2.
General transfer principles take more than just RCFs and Łoś ultraproduct theorem; we usually formulate them in terms of superstructures/universes. Most textbooks on Nonstandard Analysis will introduce the required notions (universes/superstructures), and state and prove the Transfer Principle. You could consult e.g.
Goldblatt's Lectures on the Hyperreals Part 3, or
Loeb's Nonstandard Analysis for the Working Mathematician Chapter 2.
3.
All fields satisfying your condition have cardinality at least continuum. This follows from the fact that every Dedekind cut $c \subseteq \mathbb{R}$ defines a function $f_c: \mathbb{R} \rightarrow \{0,1\}$, and you can distinguish two different cuts $c \subseteq d$ using the first-order theory of the structure $(\mathbb{R}, +, -, \cdot, f_c, f_d, 0, 1)$ simply by finding a rational $\frac{p}{q}$ that belongs to $d$ but not to $c$. This is possible because $\mathbb{Q}$ is dense in $\mathbb{R}$.
Since $\mathfrak{R} \models \exists! x_c. f_c(x_c) = 0 \wedge (\forall y. (\exists z. y - x_c = z^2) \rightarrow f_c(y) = 1)$, and $\mathfrak{R} \models x_c \neq x_d$ for any cut $d \neq c$ and corresponding unique $x_c, x_d$, a field satisfying your property will have at least as many elements as $\mathbb{R}$. Moreover, if you take a proper subset $S \subseteq \mathbb{R}$ containing $0,1$ and closed under all the usual operations $+, -, \cdot$, then you will not be able to "extend" $f_c: \mathbb{R} \rightarrow \mathbb{R}$ to $S$ for any $c \not\in S$. So $\mathbb{R}$ is minimal with this extension property.
Best Answer
The original 1966 classic text is:
Robinson, Abraham, Non-standard analysis, Princeton, NJ: Princeton Univ. Press. xix, 293 p. (1996). ZBL0843.26012.
This is a reprint of the 1974 second edition.