eg. If we are trying to find the Net Force, why don't we find the Net acceleration, which would be something like "Centripetal acceleration - acceleration due to gravity" (at the bottom of the loop)?
Forget about the idea of having contributions to acceleration. An object has only one acceleration, not a sum of several.
- There can be many forces, and they all combine into a net force. The object is being pushed a bit from this side, a bit from that side, a bit from behind, a bit from above and all in all, all these are summed up to a net force $F_{net}$.
- But you are not accelerating a bit sideways and a bit to the other side and a bit backwards, which should then combine into one acceleration. Acceleration is not made of contributions. There can be several forces, but they together result in one acceleration - they don't cause an acceleration each which is then summed up to a "net" acceleration.
Newton's 2nd law does indicate this: The formula is not $\sum F=m\sum a$ but only $\sum F=ma$; the $F$ is a sum of many $F$'s, but the $a$ is not a sum of many $a$'s.
Therefore there is no such thing as "net" acceleration - it is just acceleration. And in the case of uniform circular motion, where this acceleration must be pointing inwards, this acceleration has been named: centripetal acceleration. It is not another "type" of acceleration - just a name we call it, when it causes a circular motion.
And for such inwards-pointing acceleration to be present, all forces acting on the object combined must cause it. At the bottom of the vertical loop, there is weight pulling down and normal force upwards, and those must together cause upwards acceleration. (You don't subtract the "acceleration contributions" from each other, you just combine the forces and see what acceleration that sum causes.)
If it wasn't upwards, it wouldn't be a uniform circular motion.
why $F_{Net}=ma=\frac{mv^2}r$
People have fond out that if a motion is uniformly circular, the acceleration is always $a=\frac{v^2}{r}$ and pointing towards the centre. This can be proven seperately and has gotten nothing to do with forces.
Newton's law always holds, $F_{net}=ma$, and so in the case of uniform circular motion, the $a$ in this law can be replaced with $a=\frac{v^2}{r}$.
So if I was to find the Net Force on 60kg person at the bottom of a roller coaster of radius 9m, travelling at 18.8 $m/s^2$, I could just use $F_{Net}=\frac{mv^2}r$? This seems to me like we just ignored gravity :/
This isn't ignoring gravity - you just aren't done. Gravity will surely enter as a part of the net force $F_{Net}$. As you even correctly wrote yourself, the net force does consist of gravity downwards and normal force upwards, $F_{Net}=F_n-mg$. So plug this into the expression and the gravity influence (the weight $mg$) is indeed included:
$$F_{Net}=\frac{mv^2}r\quad\Leftrightarrow\quad F_n-mg=\frac{mv^2}r$$