$\textbf{Definition:}$ 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?
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 $(\mathbb{Z},+)$.