Let $L$ be a first-order language, and let $F[v]$ be an $L$-formula with one free variable. Suppose that $\mathfrak{M}$ is an $L$-structure such that $\mathfrak{M} \vDash F$, that is $\{x \in M,$ satisfying $F[v] \}$ is non empty.
Now, we know that $\vdash F[v] \Leftrightarrow \forall v F[v]$, but there is no reason for the above set to be all of $M$.
Thus, we have $\mathfrak{M} \vDash F[v]$ and $\mathfrak{M} \nvDash \forall v F[v]$ even though $\vdash F[v] \Rightarrow \forall v F[v]$. This means that a model $\mathfrak{M}$ can satisfy a formula $F[v]$ without satisfying its syntactical consequences (as in, being a model is not stable by syntactic consequences). Isn't it weird? Why are we defining the deduction rules of the syntax like that?
Careful.
(1) You need to distinguish
$$\vdash F[v] \rightarrow \forall v F[v]$$
from
$$\vdash F[v] \Rightarrow\ \vdash\forall v F[v]$$
The first fails in standard FOL systems (e.g. Mendelson's)
(2) It is also unhappy to write $\mathfrak{M} \vDash F[v]$; it is $\mathfrak{M}$-plus-an-assignment that satisfies the open wff.