I'm trying to disprove: $\Gamma\vDash_{v}B\Longleftrightarrow\Gamma\cup\{\neg B\}$ isn't satisfiable by a model (for every assignment). In first order logic.
where $\vDash_{v}$ means that for every assignments in the model (M) the Logical derivation is correct.
This is contrary to : Show that $\Gamma \cup \{\neg \phi\}$ is satisfiable if and only if $\Gamma\not \models \phi$ where the the Logical derivation includes the assignment itself (usually marked by $\vDash_{t}$)
The "trick" is in the use of universal quantifier (in the meta-): every model $\mathcal M$, every assignment $v$.
"$Γ \cup \{ ¬ B \}$ isn't satisfiable by a model (for every assignment)" means that, for a specific model $\mathcal M$ we have that $\mathcal M,v \nvDash \Gamma, B$ for every $v$, where "$\Gamma \vDash_v B$" means that: for every $\mathcal M,v$, if $\mathcal M,v \vDash \Gamma$, then $\mathcal M,v \vDash B$.
We can imagine the following example: let $\mathcal M$ a model of the usual Euclidean space, let $\Gamma$ the set of geometric axioms without the parallel postulate and let $B$ the parallel postulate: we have that $\Gamma \cup \{ \lnot B \}$ is unsatisfiable in $\mathcal M$.
But $\Gamma \nvDash B$, because we know that the parallel postulate in independent from the other Euclid's axioms.
A sort of "toy model" $\mathcal M$ where the RHS holds is a model with a single-element domain: $D = \{ a \}$ and an axiom-set $\Gamma = \{ \forall x (x=a \lor x=b) \}$.
We clearly have that $\{ \forall x (x=a \lor x=b) \} \cup \{ \lnot (x=a) \}$ is unsatsfiable in $\mathcal M$.
But $\forall x (x=a \lor x=b) \nvDash (x=a)$ because we may consider another model $\mathcal M_1$ with domain $D_1 = \{ a , b \}$ where $(x=a)$ is not satisfied by every assignment.