Soundness and Completeness for a single Model Only? Question modified to hopefully answer the questions (I'm a physicist to all might not be mathematically watertight)
In Enderton "A Mathematical Introduction to Logic", logical Implication is defined as:
Logical Implication $\models$ : Let $\Gamma$ be a set of wffs (well formed functions), $\varphi$ a wff. Then $\Gamma$ logically implies $\varphi$ iff for every structure X for the language and every satisfaction function s:V $\mapsto $ |X| such that X satisfies every member of $\Gamma$ with s, X also satisfies $\varphi$ with s.
In Enderton a model is defined as :
X is a model of $\varphi$ iff X satisfies $\varphi$ with every function s:V $\mapsto $ |X|. X is a model of a set $\Sigma$ of sentences iff it is a model of every member of $\Sigma$, written as $X \models \varphi$ (Enderton uses $\models_X \varphi$, to clarify that $\models_X$ isnt the Logical implication symbol $\models$).
As a result logical implication $\models$ includes all models and not just one.
Enderton then proves Soundness and Completeness between $\models$ and $\vdash$:
$$ \Gamma \models \varphi \iff \Gamma \vdash \varphi \tag{1}$$
Question : As the Forcing Truth Lemma (taken from Kunen) looks to be effectively a form of (1) but only applies to one model (G being like a satisfaction function):
$$ \forall G \; [ M[G] \models \varphi[G] \iff \exists p \in G (p \Vdash \varphi)^M] \tag{2}$$
are there general techniques for changing $\vdash$ / $\Vdash$ into a similar model specific deductive system (call it MSDS) to make Completeness and Soundness hold for a single model?
$$ \forall s (X,\Gamma[s] \models \varphi \iff \Gamma[s] \; MSDS \; \varphi) \tag{3}$$
or maybe start with (3) and see what properties MSDS needs (and maybe slightly modify the satisfaction functions used) ? If this could be done then MSDS would only deduce what is true in the specific Model and be incapable of deducing anything else.
 A: I'm not totally sure I understand your question, but I think I'm able to make some helpful comments.
First, let's consider $\Gamma\models_X \varphi$ when $\Gamma$ and $\varphi$ have no free variables. I'll assume that the definition of this relation is: If $X\models \Gamma$, then $X\models \varphi$.
We can indeed modify $\vdash$ to a relation $\vdash_X$ so that $\Gamma\models_X \varphi$ iff $\Gamma\vdash_X \varphi$, when $\Gamma$ and $\varphi$ have no free variables. Just use the original proof system used to define $\vdash$, but add every sentence true in $X$ as an axiom!
This might not be so satisfactory, since we'd like to be able to recognize proofs, and there is no general algorithm for determining which sentences are true in an arbitrary structure $X$. If the complete theory of $X$ is computably (recursively) axiomatizable, then it suffices to add the sentences in the computable axiomatization of $X$ as axioms to our proof system, and the result is a computable proof system defining $\vdash_X$. On the other hand, there are structures, like $(\mathbb{N},+,\times)$, whose theories are not computably enumerable. For such structures, there is no hope to have a computable proof system defining $\vdash_X$, since we could computably enumerate proofs $\vdash_X \varphi$ in such a system, and thus enumerate $\mathrm{Th}(\mathbb{N},+,\times)$.
Ok, what about the case of free variables? When $\Gamma$ and $\varphi$ have free variables, I'll assume the definition of $\Gamma\models_X \varphi$ is: For all assignments $s$ from the free variables of $\Gamma$ and $\varphi$ to $X$, if $X\models \Gamma[s]$, then $X\models \varphi[s]$.
In this case, simply adding the sentences true in $X$ as axioms is not sufficient. This amounts to restricting attention to all models of $\mathrm{Th}(X)$, i.e., to all structures elementarily equivalent to $X$. This strategy works to understand entailment between sets of sentences, which will be the same for all structures elementarily equivalent to $X$. But different structures elementarily equivalent to $X$ might differ on entailments between sets of formulas with free variables.
As an example, suppose the language consists of infinitely many constant symbols $\{c_n\mid n\in \mathbb{N}\}$. Let $X$ have underlying set $\mathbb{N}$, with $c_n$ interpreted as $n$. Let $\Gamma = \{x\neq c_n\mid n>0\}$. Then $\Gamma\models_X (x = c_0)$, since $0$ is the only element of $X$ which is not equal to any $c_n$ with $n>0$. Of course, in an elementary extension $X\prec X'$, $X'$ has elements which are not equal to $c_n$ for any $n$, so $\Gamma\not\models_{X'}(x = c_0)$. So using the definition of $\vdash_X$ proposed above, we do not have $\Gamma\vdash_X (x = c_0)$.
This example also shows that $\models_X$ is not compact, and hence  there is no finitary proof system giving $\Gamma\models_X\varphi$ iff $\Gamma\vdash_X \varphi$. Explicitly: there is no finite subset $\Delta\subset_{\mathrm{fin}} \Gamma$ such that $\Delta\models_X(x = c_0)$. In fact, there is no proper subset $\Delta\subsetneq \Gamma$ such that $\Delta\models_x(x = c_0)$. A proof of $(x = c_0)$ from $\Gamma$ must use every formula in $\Gamma$, and hence must be infinite.
