First, consider the two boxes as a single system with total mass $M = m_1 + m_2$. The net external force on the system is the applied force minus the total friction, $F_{net} = F - f_1 - f_2$.

Applying Newton's second law to the system gives the common acceleration $a$:

$$F - f_1 - f_2 = (m_1 + m_2)a$$ $$a = \frac{F - f_1 - f_2}{m_1 + m_2}$$

Next, isolate box $m_2$. The forces acting on it in the horizontal direction are the contact force from box 1, $F_{12}$, and the frictional force $f_2$. Applying Newton's second law to box $m_2$:

$$F_{12} - f_2 = m_2 a$$

Solving for $F_{12}$ and substituting the expression for $a$:

$$F_{12} = m_2 a + f_2 = m_2 \left( \frac{F - f_1 - f_2}{m_1 + m_2} \right) + f_2$$

To simplify, find a common denominator:

$$F_{12} = \frac{m_2(F - f_1 - f_2) + f_2(m_1 + m_2)}{m_1 + m_2}$$ $$F_{12} = \frac{m_2 F - m_2 f_1 - m_2 f_2 + m_1 f_2 + m_2 f_2}{m_1 + m_2}$$ $$F_{12} = \frac{m_2 F - m_2 f_1 + m_1 f_2}{m_1 + m_2}$$