The proof to
$\Gamma \models \phi [\nu /\kappa ] \: \Rightarrow \: \Gamma \models \left ( \forall \nu \right )\phi$
(Provided that no sentences in $\Gamma\:$ have occurrences of $\kappa$ and also that $\kappa$ does not occur in $\phi$).
Goes as follows:
Suppose that $\:\Gamma \models \phi [\nu /\kappa ]$. For RAA, assume that it is not the case that $\:\Gamma \models \left ( \forall \nu \right )\phi$. Then there is some interpretation $\:\mathbf{I}$ that satisfies $\:\Gamma$ but not $\:\left ( \forall \nu \right )\phi$. Since $\kappa$ does not occur in $\phi$, there is some variant $\textbf{I}_{\kappa}$ of $\:\mathbf{I}\:$ with respect to $\kappa$ such that $\phi [\nu /\kappa ]$ is false under $\textbf{I}_{\kappa}$. But also $\kappa$ does not occur in any sentence of $\:\Gamma$. Since the only difference between $\:\mathbf{I}$ and $\textbf{I}_{\kappa}$ is what they assign to $\kappa$, and since $\:\mathbf{I}\:$ satisfies $\:\Gamma$, therefore $\textbf{I}_{\kappa}$ satisfies $\:\Gamma$. Thus, $\textbf{I}_{\kappa}$ satisfies $\:\Gamma$ but not $\phi [\nu /\kappa ]$, which contradicts the hypothesis that $\Gamma \models \phi [\nu /\kappa ]$. $\mathbf{Q.E.D.}$
My question is,
Why does it not matter what $\kappa$ is chosen for $\:\Gamma$? Specifically, I am not understanding the significance of the sentence,
Since the only difference between $\:\mathbf{I}$ and $\textbf{I}_{\kappa}$ is what they assign to $\kappa$, and since $\:\mathbf{I}\:$ satisfies $\:\Gamma$, therefore $\textbf{I}_{\kappa}$ satisfies $\:\Gamma$.
Does $\:\mathbf{I}\:$ satisfying $\:\Gamma$ implicitly mean that any arbitrary constant that does not occur in $\Gamma$ satisfies all substitution instances inside $\Gamma$? Nothing suggests that $\Gamma$ is satisfied for all interpretations so why doesn't any $\kappa$ affect it?
Let's fix a first order language $\mathcal{L}$, where $\Gamma$ is a set of sentences in the language $\mathcal{L}$. Suppose $I$ is a model of $\Gamma$ satisfying $\lnot \forall v \phi(v)$. That is, there is $a \in I$ such that $I \models \lnot \phi(a)$.
If we expand the language to $\mathcal{L} \cup \{ \kappa \}$, where we add a new constant symbol, $\Gamma$ is still a set of sentences in this language. In order to make $I$ to be a model in this expansion, we need an interpretation of the constant symbol $\kappa$. So let's call $I_\kappa$ the interpretation $I$ with $\kappa$ interpreted as the element $a$. $I_\kappa$ satisfies the same $\mathcal{L}$-sentences as $I$, since it has the same universe and the same interpretations for any symbols in $\mathcal{L}$. In particular, if $I \models \Gamma$ then $I_\kappa \models \Gamma$.