The relationship between the force on the charge ($\vec{F}{\text{charge}}$ or $\vec{F}{\text{Coulomb}}$) and the total force on the field ($\vec{F}{\text{field}}$ or $\vec{F}{\text{surface}}$) for a static charge configuration is that they are equal in magnitude and direction.
$$ \vec{F}{\text{surface}} = \vec{F}{\text{Coulomb}} $$
In this verification, the surface force on the field ($\vec{F}_{\text{surface}}$) is calculated for the region $x_3 < 0$, which contains the lower charge.
This relationship is derived from two fundamental principles that must hold for static equilibrium:
Static Equilibrium of the Field ($\sum \vec{F}_{\text{field}} = 0$):
The total force on the electromagnetic field in the lower region ($x_3 < 0$) must be zero. This total force is the sum of the force the charge exerts on the field ($\vec{F}{q \to \text{field}}$) and the force the upper field exerts on the lower field across the surface, which is the computed surface force ($\vec{F}{\text{surface}}$).
$$ \vec{F}{q \to \text{field}} + \vec{F}{\text{surface}} = 0 \implies \vec{F}{\text{surface}} = - \vec{F}{q \to \text{field}} $$
Newton's Third Law (Action-Reaction):
The force the charge exerts on the field ($\vec{F}{q \to \text{field}}$) is equal in magnitude and opposite in direction to the force the field exerts on the charge ($\vec{F}{\text{Coulomb}}$).
$$ \vec{F}{q \to \text{field}} = - \vec{F}{\text{Coulomb}} $$
Conclusion:
Substituting the second equation into the first yields the final relationship:
$$ \vec{F}{\text{surface}} = - (-\vec{F}{\text{Coulomb}}) = \vec{F}_{\text{Coulomb}} $$
In the case of the two equal charges ($q$ and $q$ separated by $2d$):
$$ \vec{F}{\text{surface}} = \vec{F}{\text{Coulomb}} = -\frac{q^2}{4\pi\epsilon_0 (2d)^2} \vec{e}_3 $$
Both forces point downward in the $-\vec{e}_3$ direction.