Why do thin films need to be thin?

You are completely right in stating that the same effect should occur for thicker slabs. There are however at least 3 practical reasons why the effect is more easily observed in thin films.

1. Light sources are typically not completely monochromatic. They emit slightly different colours at the same time. Imagine a light source whose wavelength varies by 0.1 percent. For a layer which is a couple of wavelengths thick, all colours will interfere destructively under the same angle. However when the layer is 1000 wavelengths thick, one colour will interfere constructively, while the other interferes destructively. The interference pattern will thus be lost. BTW, the length over which a light source can interfere is called the coherence length.

2. Light sources are typically not infinitesimally small. Instead of being a point source, the source has a certain width. This means that the light effectively arrives under various angles at a certain point. For thin samples this does not matter as all these angles will interfere constructively at the same time. However for a thick layer, light under certain angles will interfere constructively, while other angles interfere destructively and the interference pattern is lost.

3. Light sources are typically not completely spatially coherent. This means that the wave front is not completely neat and flat. The distance over which the wavefront is still flat (~up to half a wavelength) is called the coherence width. The result is that when a part of the beam interferes with another part of the beam, which is further away than the coherence width, the interference pattern is lost. For a thick layer, when light is applied under an angle, the light typically interferes with another part of the beam further than the coherence width. Note that mathematically reasons 2 & 3 are actually the same.

4. It is difficult to make thick flat films. However not as impossible as it might seem. Wafers are typically polished up to atomic flatness over micrometer distances and vary only hundreds of nanometers over millimeter distances.

Ignoring for simplicity's sake the usual refraction that takes place at the interfaces of the media, if:

$$|OA|+|AB|=D\big(\frac{1}{\cos \theta}+\tan \theta\big)=n\lambda$$

with $n$ an integer and $\lambda$ the wave length, then we have positive interference.

But as we increase $D$, the distance $\Delta$ also increases, so for high values of $D$ these rays can no longer interfere.