Intuition behind the center of mass I know how the center of mass is defined, mathematically. It is the mass weighted average position of all the particles of a system. But calculating centers of mass and solving kinematic and dynamics problems related to them has only been plugging and chugging formulas so far, with no physical insight on what im actually doing
There must be a way to intuitively interpret this. What does the mass weighted average position mean, exactly, and what does it imply? Secondly, why is this mass weighted position the (only) point always behaving as a point particle with mass equivalent to the total mass in regards to newton's laws. The mathematical proof of this is, again, very accessible to me, but this surprising result is surely no coincidence and must have some physical reasoning.
I’m afraid that my knowledge of statistics is not very strong so please go a little easy with statistics.
 A: Consider a rigid body as a collection of particles moving together. The center of mass is the unique point in space where the following is true
$$ \sum_i m_i \boldsymbol{r}_i = \left( \sum_i m_i \right) \boldsymbol{r}_{\rm COM} $$
But that does not look very intuitive, unless you take the time derivative and figure out the total momentum of the system
$$ \boldsymbol{p} = \sum_i m_i \boldsymbol{v}_i = \left( \sum_i m_i \right) \boldsymbol{v}_{\rm COM} $$
The intuition behind the center of mass, is the point in space whose velocity can be factored out the total momentum expression.
This is quite powerful as it leads to Newton's 2nd law, again by taking the derivative
$$ \sum \boldsymbol{F}_i = \tfrac{\rm d}{{\rm d}t} \boldsymbol{p} = \left( \sum_i m_i \right) \tfrac{\rm d}{{\rm d}t} \boldsymbol{v}_{\rm COM} $$
$$ \boldsymbol{F} = m \,\boldsymbol{a}_{\rm COM} $$
A: Consider an object made  of 5 points of  masses $M_1$, $M_2$, $M_3$, $M_4$, $M_5$. Now imagine that I place the object in midair (which is not possible, but imagine for a moment). Since the object is on earth, it experiences a gravitational force, and the force on the particle with mass $M_1$ is $M_1g$.
I try to stop it with an equal and opposite force ($-M_1 g$) with my finger, so the net force on this particle is $0$. Don't forget that there are 4 more particles, so in the pursuit of not allowing the object to fall onto the ground, I give five equal and opposite forces on those five points, namely $M_1g$, $M_2g$, $M_3g$, $M_4g$ and $M_5g$.
Here we can say that the total gravitational force on the object is $F=M_1g+M_2g+M_3g+M_4g+M_5g$. The force given by me (pushing against the object) must be $-F$ to prevent the object from falling. This force $-F$ cancels the gravitational force $F$, since $F+(-F)=0$. So in order to stop it from falling, I must keep my 5 fingers on the 5 point masses.
Now is it possible to exert the same force on a single point and keep it from falling under the influence of gravity? Well, by trial and error, I find that (consider the object to be symmetrical) if I apply the force $F$ on the geometric centre, the object does not fall, now this has a mass right?
Consider the mass of the point to be $K$, the net gravitational force would be $F$, so to counteract the gravitational force, I must apply an equal and opposite force $-F$. Let the force be $Kg$. Both the forces have an equal magnitude, i.e. $Kg=M_1g+M_2g+M_3g+M_4g+M_5g$, on eliminating $g$ we get $K=M_1+M_2+M_3+M_4+M_5$.
A mathematical definition would be: A centre of mass is the point where, when we balance using a single finger, the net torque is zero.
A: The logic offered by user794763 is basically correct.  If you exert a force which acts along a line through the center of mass of a solid object, it will not cause a rotation.
