Intuitive understanding of why the sum of nth roots of unity is $0$

Consider the cube roots of unity. You can write them in the plane with $x$ being the real part and $y$ being the imaginary part. They give three vectors all with unit length, equally spaced around the unit circle. Now let's add the two vectors that are not 1 (ie, the two that face to the left) and you get $-1$ as shown in the second picture below (the light vectors are gone and their sum is the left-pointing vector which represents $-1$). Of course you can add any two vectors you like, but it will always give you a vector that is the negative of the remaining one. Adding the final two will give 0. This happens for any number of vectors when they are a complete set of all roots of unity.

Note: This is far from a rigorous proof, but I'm trying to appeal to your intuition. The picture isn't nearly so nice for 5-ths roots of unity.

To address your edited question, yes, $2\cos(72) + 2\cos(144) = -1$. It's not obvious, so I don't think you're being silly. It comes from the general formula

$$ \sum_{k=1}^n \cos \frac{2 \pi k}{n} = 0 $$

which, with a tiny amount of manipulation, gives you the formula above. This identity gives another proof that the $n$-th roots of unity sum to 0: the $\sin$ values will cancel each other out leaving a cosine sum equivalent to the above. The normal proof for this formula uses complex numbers, but you can also obtain a more elementary proof using trig identities. However, I would argue that your request for an intuitive understanding is better served by the respondents using "center of mass" type arguments than this trigonometry stuff.

The $n$ roots of unity of order $n$ (all of them, not just the primitive ones) are equally spaced around the unit circle. If their sum was nonzero, it would have an argument (angle relative to the $x$ axis) which would be a violation of symmetry. Therefore, by symmetry, the sum must be 0.

Appealing to physical intuition, it is just the center of mass.

Think of the roots of unity as the coordinates of $n$ identical balls equally spaced around a circle in $\mathbb{R}^2$. What is the location of the center of mass? It is the center of the circle. Consequently, adding the roots of unity gives the center of the unit circle, which is $0$.