Why is Bragg diffraction small-angle for thicker layers and/or larger molecules or unit cells? And wide-angle for small or shallow crystals? It is easy to find equations that quantify this on the internet, but not an explanation as to why...
Also, does this apply to electrons and neutrons?  Or just X-ray reflection/diffraction?
 A: It has to do with optical laws, regardless of the wave you use.
Bragg's law gives you the angles in which there is a maximum, but it doesn't say anything about the intensity you will detect on them.
If you perform the calculations, you'll find out that the intensity of the maxima rises with $N^2$, but the width decreases as $1/N$.
You can check this in a simple 1D model. Consider a chain of $N$ equally spaced atoms at a distance $a$. They're indexed from $0$ to $N-1$. Any lattice vector is $R=n\cdot a$
The intensity would be the sum $S$ squared.
$$S=\sum_{n=0}^{N-1} e^{-i \Delta k\cdot R}=\sum_{n=0}^{N-1} e^{-i \Delta k\cdot (na)}=\sum_{n=0}^{N-1} \left(e^{-i \Delta k\cdot a}\right)^n=\dfrac{e^{-i \Delta k\cdot Na}-1}{e^{-i \Delta k\cdot a}-1}$$
$$S=e^{-i \Delta k\cdot (N-1)a/2} \cdot \dfrac{\sin(N\Delta k a/2)}{\sin(\Delta k a/2)}$$
Taylor to First order yields
$$I\propto |S|^2 = 1\cdot \left| \frac{N k a/2}{k a /2 } \right|^2 \propto N^2$$
On the otehr hand, the distance between the two first minima is proportional to $1/N$, so the width of the peak decreasses with $1/N$.
