How is gravity consistent when you split an object into multiple pieces? Here's the picture:

In picture 1: Assume that we have 2 objects, namely A, and B. Object A with mass $2M$, and object B with mass $m$; and their distance is $r$. Then the gravitational force of A acting on B ($\vec{v}$)'s magnitude will be: $$G\times\frac{2Mm}{r^2}.$$
In picture 2: Now divide object A into 2 equal parts of mass $M$ each. The distance from the centroid of each part of A to B is $\dfrac{r}{\cos \alpha}$. The gravitational forces of 2 parts of A acting on B ($\vec{v}_1; \vec{v}_2$)'s magnitudes are: $$G\times \frac{Mm\cos^2\alpha}{r^2}.$$
Now, I'm pretty sure that if I take the sum $\vec{v}_1 + \vec{v}_2$, I wouldn't get $\vec{v}$. The direction is the same, but the magnitude isn't. They are off a factor of $\cos ^ 3 \alpha$. :(
What's going on here?
 A: In your picture 1, the gravitational force is calculated incorrectly. The formula you have used only applies between pointlike masses. You have to divide the object into elements, calculate the contributions of each element and sum up. The picture 2 is only the first step in the whole process, so actually not even your picture 2 is generally correct.
A: The problem is in your assumption that the force is $F = 2GMm/r^2$. This is true for the force on a point mass from a sphere or another point mass, but not otherwise. What you need to do is sum
each the force on each particle from every other particle. For a continuum object, $$\vec F = \int \rho \vec g\, dV$$ where $\rho$ is the density and $\vec g$ the acceleration due to gravity. Both can vary over the volume. To find $\vec g$ you sum up the contribution from all points,
$$\vec g(\mathbf x) =  G\int\rho  \frac{\mathbf y -\mathbf x}{|\mathbf x -\mathbf y|^3}\, dV$$
do as you can see it is significantly more difficult for bodies other than point masses (and spheres, it turns out). But from these expressions you can see that forces on parts of rigid bodies add, and so do the forces from parts of bodies. 
Often you can pretend that you are dealing with point masses because from far away, anything looks like a point mass. But as you have discovered, this is an approximation and not exact. 
A: The "intuitive" expectation is that both methods should give the same answer - and they do, if you do it right.
Were method 2 goes wrong is in calculating the force at distance $$d = \frac{r}{\cos{ \alpha}}$$  although this gives you the force between M and m, this is the wrong force.  The force that is required, is the force along a line between m and the center of mass of the two Ms.  This means that the distance to be used is $d {\cos{\alpha}}$, which is r! So the effective force between M and m is $$F1 = F2 = \frac{MmG}{r^2}$$ Since the two vectors have the same magnitude and direction, the resultant is$$ F1 + F2 =\frac{(2)MmG}{r^2}$$ which is the same as the force between 2M and m $$F =\frac{(2M)mG}{r^2}$$        
