why mass is additive; i.e. the transnational kinematics of some body can be determined from considering it as a single body with a total mass equal to the sum of the individual masses, acting at a point which we call the centre of mass.
The fact that you can consider a collection of particles as one large particle follows from Newton's third law. Taylor gives a nice proof of this in his Classical Mechanics book. I'll redo it here:
Say you have a collection of particles with positions $\{x_i\}$ and masses $\{m_i\}$. They are each being acted on by outside forces, $\{F_i\}$. They also interact among themselves: the force of the jth particle on the ith particle is labelled as $\{F_{ij}\}$. These forces can be thought of as the forces that hold our collection together; each particle might be attracted to every other particle, for example. Then we can consider the acceleration of the center of mass, $X$:
$$
\ddot{X}=\frac{1}{M}\sum_i{m_i\ddot{x_i}}
$$
This follows directly from the definition of center of mass: $X=\frac 1 M \sum_im_i x_i$, where $M$ is the total mass. Now, we can use Newton's second law to rewrite each $m_i \ddot{x_i}$ as the total force on the ith particle, $F_i+\sum_jF_{ij}$
$$
\ddot{X}=\frac{1}{M}\sum_i{F_i+\sum_jF_{ij}}=\frac{1}{M}\sum_i F_i +\frac{1}{M}\sum_{ij}F_{ij}
$$
Now, from Newton's third Law, $\sum_{ij} F_{ij}=0$. This is because for every term in the sum, say $F_{ab}$, there is another term $F_{ba}$, and Newton's third law says $F_{ab}=-F_{ba}$. Thus, our formula becomes
$$
\ddot{X}=\frac{1}{M}\sum_i F_i
$$
or
$$
M\ddot{X}=F_{total}
$$
From this proof, you can see the reason why we can treat extended objects as if they were single particles of larger mass is due to Newton's third law, which allowed us to cancel the sum over $F_{ij}$. Intuitively, this happens because Newton's third law says that no object, not even an extended object, can exert a force on itself. Thus, a collection of particles only accelerates due to external forces.
If you liked this proof, Taylor has a similar one for angular momentum that you might find illuminating.
I find it even more perplexing that the moment of inertia of some compound object about a given axis can be summed by finding the sum of the individual moments of inertia! This is particularly puzzling for me because the moment of inertia is proportional to the distance squared (although perhaps this has nothing to do with the problem!)
This just follows directly from the definition of $I$, as SammyGerbil pointed out.
