Is the force due to gravity between two objects immersed in a fluid correctly given by$-G(M_1-m_1)(M_2-m_2)/r^2$? ($m$ = mass of displaced fluid)

I think there is a lot of confusion here. First of all, by Newtonian gravity, the gravitational force between two distinct objects depends only on their masses and the distance between them. That's it. There are GR corrections of course but we can consider those negligible for now.

Things already go a bit wrong for me in 2). From the question it seems like you are talking about the gravitational influence of one spherical object on another object, but then in 2) you're talking about the fluid, which has no influence on the gravitational influence between the two objects. Does the fluid exert gravitational influence? If it has energy of course it does, but it in no way affects the gravitational influence between the two spherical objects.

Consider this, you are essentially talking about 3 objects here: Spheres A and B, and the fluid C. There are three gravitational forces here, $$F_{AB}$$, $$F_{AC}$$, and $$F_{BC}$$. Your question asks about $$F_{AB}$$, which as I said must be $$F_{AB} = \frac{G m_A m_B}{r^2}$$, regardless of the fluid.

Now the gravitational force from the fluid does depend on its density and distribution, but it in no way relates to the equation proposed. Let me know if you have any more questions.

The formula $$F = -G ~ \frac{\Delta M_1\Delta M_2}{r^2}$$ is correct. The force is the derivative of the energy with respect to the separation $$r$$. By displacing one of the immersed masses, say $$M_1$$, effectively only a mass $$M_1-m_1$$ is displaced. Indeed a volume of fluid with mass $$m_1$$ is displaced oppositely.

A very similar line of reasoning based on potential energy goes as follows. When a massive objects is added to the system, a mass $$m_1$$ is removed so effectively the mass added is $$\Delta M_1$$. Therefore an additional potential $$V_1=-G\frac{\Delta M_1}{r}$$ is added. Adding another object $$M_2$$ only adds mass $$\Delta M_2$$ which feels $$V_1$$, resulting in the potential energy $$V_{12} = -G ~ \frac{\Delta M_1\Delta M_2}{r}$$