Please note I am not asking how to prove the theorem, but rather, asking about a particular point on it.
When showing the first implication of the theorem we need to show that $X'''=J_{X'}(X')$; i.e. assuming that $X$ is reflexive show that $X'$ is reflexive.
Taking $\rho\in X'''$ we show that there exists an $f\in X'$ for which $J_{X'}f=\rho$. By definition of $\rho\in X'''$ we know that $\rho$ is a bounded operator. The composition $\rho\,\circ J_{X}x$ is then a bounded map from $X$ to $\mathbb F$ so that this composition is in $X'$. Set $f:=\rho\,\circ J_{X}x$. Since we have assumed that $X$ is reflexive, for all $\psi\in X''$ there exists an $x\in X$ for which $\psi=J_Xx$.
At this point we now have everything we need to show that $X'''=J_{X'}(X')$. In my notes we begin with the following equality from which follows a whole chain of equalities.
$$((J_{X'}f))(\psi)=\psi(f)$$
First consider the object on the left of the equality sign. We have $f\in X'$ so that $J_{X'}f\in X'''$. We also have that $\psi\in X''$ so that $((J_{X'}f))(\psi)$ is just a number in our scalar field $\mathbb F$. Now consider that on the right hand side of the equality. We have $\psi \in X''$ for which $\psi( f)$ is also in the scalar field $\mathbb F$. Now here is my question; on both sides of the equality we have numbers in our scalar field, but how do we know that they coincide? How do we know that applying $((J_{X'}f))\in X'''$ to $\psi\in X''$ gives us the same number as applying $\psi\in X''$ to $f\in X'$? It kind of makes sense in my head but I'm not entirely convinced.
This is just the definition of $J_{X'}$. It maps $f \in X'$ to an element of $X'''$, which acts via $X'' \ni \varphi \mapsto \varphi(f) \in \mathbb F$. Hence, $(J_{X'}f)(\varphi) = \varphi(f)$.