From the momentum equation of fluid mechanics: $${\frac{\partial}{\partial t} \int_{V_f(t)} \rho \vec{v} \text{d}\text{V}} = \int_{\sum_f(t)} \vec{\bar{\bar{\tau}}} \cdot \vec{n} \text{d}\sigma + \int_{V_f(t)}\rho \vec{f_m} \text{dV}$$

where $V_f$ and $\sum_f$ are the fluid volume and surface.

This equation means that the rate of momentum of the fluid volume equals the surface and volume forces on the system: the fluid volume. Now, if we take a control volume, use Reynolds transport theorem on the left and we express the stress tensor $\bar{\bar{\tau}}$ in a different way on the right it would lead to the following equation: $${\frac{\partial}{\partial t} \int_{V_c(t)} \rho \vec{v} \text{d}\text{V}} + \int_{\sum_c(t)}\rho\vec{v}\text{(}\vec{v}-\vec{v_c})\cdot\vec{n}d\sigma = -\int_{\sum_c(t)} p \vec{n} \text{d}\sigma + \int_{\sum_c(t)} \vec{\bar{\bar{\tau'}}}\cdot\vec{n}d\sigma + \int_{V_c(t)}\rho \vec{f_m} \text{dV}$$ where Vc and $\sum_c$ are the volume and the surface of the control volume. $\sum_c = \sum_f + \sum_1 + \sum_2$, where $\sum_1$ and $\sum_2$ are the surfaces needed to close the control volume.

I know that the surface and volume of the control volume are chosen to be equal to the system's volume at that instant. What I don't understand is the correspondence between the forces of the control volume and the fluid. You see, if you take a control volume you are studying the forces on the fluid surface plus the forces on the boundaries that are left. That would mean that the force $F_c = F_f + F_1 + F_2$, where $F_1$ and $F_2$ are the forces on the surfaces that close the control volume. But, according to momentum equation $F_c = F_f$, right? This is what doesn't make sense to me.