Can we add attraction forces? From Newton's Law of gravitation we know that:
$$F=G\frac{m_1m_2}{d^2}$$
For simplicity, let's say that both $m$ are $1\;\mathrm{kg}$ and that the distance apart was $1\;\mathrm{m}$. Yielding $G$ as the attraction force in Newtons. Hence $F= 6.67\times 10^{-11}\;\mathrm{N}$. 
Now what if you had two $0.5\;\mathrm{kg}$ glued to each other and there was another two $0.5\;\mathrm{kg}$ glued together $1\;\mathrm{m}$ apart. If we were to find the gravitational attraction force of two $0.5\;\mathrm{kg}$ masses which are $1\;\mathrm{m}$ apart and multiply it by two since we have two of them that would yield a different answer than treating the two $0.5\;\mathrm{kg}$ that are glued together as one entity. Why does it yield a different answer? 
I observed this same phenomenon with Coulomb's law. (Serway & Faughn)
Suppose that $1.00\;\mathrm{g}$ of hydrogen is separated into electrons and protrons. Suppose also that the protons are placed at the Earth's north pole and the electrons are placed at the south pole. What is the resulting compression force on the Earth? If I was to find the attraction force of one proton and electron than multiply it by Avogadro  constant that would yield a different answer than saying that the charge of each particle is the elemental charge times Avogadro 's constant.
So which is the proper way to do it? 
 A: You need to multiply by four not by two. To see why let's draw the situation:

You are assuming the situation is as shown in the top diagram. So the two $M_1$s attract each other and the two $M_2$s attract each other. Those are the forces shown by the red lines. 
But you also need to include the force between $M_1$ on one side and $M_2$ on the other. Those are the green lines in the second diagram. So the total force will be:
$$ F = F_{1-1} + F_{2-2} + F_{1-2} + F_{2-1} $$
Suppose we make the mass of each of the balls $M$, so if we combine them the total mass on each side is $2M$. If we combine the masses first then calculate the force we get:
$$ F = \frac{G(2M)(2M)}{d^2} = 4\frac{GM^2}{d^2} $$
If we calculate the forces treating the masses separately we get${}^1$:
$$\begin{align}
 F &= F_{1-1} + F_{2-2} + F_{1-2} + F_{2-1} \\
   &= \frac{GMM}{d^2} + \frac{GMM}{d^2} + \frac{GMM}{d^2} + \frac{GMM}{d^2} \\
   &= 4\frac{GM^2}{d^2} 
\end{align}$$
So the forces are the same.
This also applies to the example of the electrostatic force that you mention.

${}^1$ Strictly speaking the distances $M_1 \rightarrow M_2'$ and $M_2 \rightarrow M_1'$ are slightly greater than the distance $M_1 \rightarrow M_1'$ and $M_2 \rightarrow M_2'$ but we'll assume this difference is negligibly small.
A: If we consider the two masses of $1(kg)$ as point particles, then the force between them when they are $1(m)$ separated is $G(N)$. 
The force between two point particles of $0,5(kg)$, who are also separated $1(m)$, is $0,25G(N)$. Now if we glue a point particle with mass $0,5(kg)$ to each of these two particles, the masses of the two $0,5(kg)$ particles become $1(kg)$ again, and thus the force between them is $G(N)$ again, so there is no difference.
