Why can any type be realized? I couldn't find this question asked previously, which means it's probably an especially daft question.
Given an $\mathcal{L}$-structure $\mathcal{M}$, my textbook defines an $n$-type over $A\subseteq M$ to be a set $p$ of sentences all in the same $n$ free variables such that $p\cup Th_A(\mathcal{M})$ is satisfiable. ($Th_A(\mathcal{M})$ is here the complete theory of $\mathcal{M}$ considered as a structure in the language $\mathcal{L}\cup\{c_a: a\in A\}$.)
The proof in the book proceeds by showing that since the union of $p$ and the elementary diagram of $\mathcal{M}$ is satisfiable, there's a model $\mathcal{N}$ into which $\mathcal{M}$ is elementarily embedded and which obviously satisfies $p\cup Th_A(\mathcal{M})$. This much I understand perfectly. 
The (imo, important) step of showing that there's an $\overline{a}\in N^n$ such that it satisfies every formula in $p$ is sort of brushed over. "Now let $c_i\in N$ be the interpretations of $v_i$. Then $(c_1,\ldots,c_n)$ is a realization of $p$." (This is David Marker; Chang & Keisler are even less helpful.)
I have clearly misunderstood something important; I know what the interpretation of $v_i$ with respect to a sequence $\overline{a}\in N^m$ for $m > i$ is, but I don't see a warrant in the proof or in the definitions surrouding interpretation for such a thing as "the" intepretation of a free variable. Without such a thing,  though, I'm not sure what actually guarantees realization of $p$.
So, what's the step I'm missing here?
 A: It all just hinges on the definition of satisfiable formula. Note that you misquoted Marker (you wrote sentence instead).
There is of course only one sensible possible definition (which I didn't manage to locate in Marker, unfortunately):

A set of formulae $\Phi$ in the free variables $(x_n)_n$ is satisfiable if there exist a model $\mathcal M$ and elements $(m_n)_n \in M$ such that for each $\phi \in \Phi$: $$\mathcal M \models \phi(m_1,\ldots,m_n)$$ (where $n$ is the highest index of a variable occurring in $\phi$).

Now the desired existence of $\bar a$ is an immediate consequence of the definition.
A: The definition as stated is a bit sloppy. In order to make it more precise, take new constants $c_{v_1},...,c_{v_n}$ representing the variables $v_1,..v_n$. So $p\cup Th_A(M)$ is consistent  really means $\{\phi(c_{v_1},...,c_{v_n}): \phi(v_1,...,v_n)\in p\}\cup Th_A(M)$ is consistent in the language $L'=L\cup\{c_a: a\in A\}\cup\{c_{v_1},...c_{v_n}\}$.
A: Since you understand that $\mathcal N$ satisfies $p\cup Th_A(\mathcal M)$, there must be at least one value assignment for the free variables in $p$ such that every formula in $p$ is true under this value assignment. That's what satisfiability means.
The sentence you quote then supposed that you choose one such value assignment and define the $c_i$s as the values the variables take in that particular value assignment. Since you're proving an existential statement, you don't have to come up with a unique $\bar a$ that satisfies every formula in $p$ -- it is enough to show that at least some $\bar a$ exists here.
