From the mental picture you have drawn (very well) in your question, the block $1$ with mass $m_2$ sits atop of the friction surface (coefficient $\mu_2$) of the block $2$ with mass $m_1$, which itself lies on a friction surface (coefficient $\mu_1$). The force $\vec{F} = F \cos \theta \hat{i} + F \sin \theta \hat{k}$ is applied to the block with mass $m_1$, where $0 \leq \theta$ and the unit vectors $\hat{i}$ and $\hat{k}$ indicate the horizontal and vertically upwards directions respectively. Finally a rope (with tension say $T$) is attached to the block $2$ preventing it from moving in the direction of the force $(\vec{F} \cdot \hat{i}) \hat{i}$.
Let $N_1$ be the magnitude of the normal reaction force acted by the friction surface on block $1$ and $N_2$ be the magnitude of the normal reaction force acted on the block $2$ by th esurface of block $1$. From the free body diagram, the static equations of motion for the $\hat{k}$ direction will read, $$0 = N_1 - (N_2 + m_1 g - F \sin \theta),$$ $$0 = N_2 - m_2 g,$$ for the blocks $1$ and $2$ respectively, due to the fact that neither block accelerates in the vertical direction. The static equation of motion for the block $2$ in the $\hat{i}$ direction is $$0 = T - \mu_2 N_2,$$ since the block will not (presumably) accelerate horizontally, while the dynamic equation of motion for the block $1$ in $\hat{i}$ direction is $$m_1 a_{1 \hat{i}} = F \cos \theta - \mu_1 N_1 - \mu_2 N_2,$$ where $a_{1 \hat{i}}$ is the horizontal acceleration of the block $1$. These calculations can be directly by doing the algebra, to explicitly find the magnitude of the force $\vec{F}$ such that $a_{1 \hat{i}} > 0$. Indeed, the required constraint is given as $$F > \frac{(\mu_1(m_1 + m_2) g + \mu_2 m_2 g)}{\cos \theta + \mu_1 \sin \theta}.$$
