how can we show directly by the "definition" that free groups are not pseudofinite? $\textbf{Definition:}$
 An $\mathcal{L}$-structure $\mathcal{M}$ is pseudofinite if for all $\mathcal{L}$-sentences $\sigma$, $\mathcal{M}\models σ$ implies
that there is a finite $\mathcal{M_{0}}$ such that $\mathcal{M_{0}} \models \sigma$.
Although, there exists a proof that says the centralizers of elements in a free group are definable and isomorphic to $(\mathbb{Z}, + )$, so free groups are not pseudofinite. (Assuming we know that $(\mathbb{Z}, + )$ is not pseudofinite group), how can we show directly by the "definition" that free groups are not pseudofinite?   
 A: Well, to spell out the strategy outlined in the proof you refer to, first one observes that $(\mathbb{Z},+)$ is not pseudofinite. One way to do this is to exhibit a definable function which is injective but not surjective, like the doubling function $f(x) = x + x$. So for example $$\left(\forall x\, \forall y\, (x + x = y+y \rightarrow x = y)\right)\land \left(\exists x\forall y\, (y+y \neq x)\right)$$
is true in $\mathbb{Z}$ but not in any finite group. 
Now any non-trivial free group $(F,\cdot)$ interprets $(\mathbb{Z},+)$ as the centralizer of any non-identity element. So we just need to translate the above sentence through this interpretation.  For example, $$(\forall x\, \forall y\, (x\cdot x = y \cdot y \rightarrow x = y)) \land \exists z\, \exists x\, ((x\cdot z = z \cdot x) \land \forall y ((y\cdot z = z\cdot y)\rightarrow (y \cdot y \neq x)))$$
is true in $F$ but not in any finite group. 
Explicitly, the sentence says that the squaring map $f(x) = x\cdot x$ is an injective function from the group to itself (and since if $x$ commutes with $z$, then $x\cdot x$ automatically commutes with $z$, this implies that $f$ restricts to an injective function $C(z)\to C(z)$ for any element $z$), and there is some element $z$ such that $f\colon C(z)\to C(z)$ is not surjective. 
Edit: After reflecting on this, you may notice that the interpretation argument is a bit more complicated than necessary. In fact, the squaring map $f(x) = x\cdot x$ is already an injective but not surjective function $F\to F$, so we can use the sentence $$\left(\forall x\, \forall y\, (x \cdot x = y\cdot y \rightarrow x = y)\right)\land \left(\exists x\forall y\, (y\cdot y \neq x)\right)$$ just like in $(\mathbb{Z},+)$.
