why do we get a zero coefficient for frequencies not in the sound-wave using Fourier transform?

A frequency $f$ being present in the signal $g(t)$ means that we can decompose $g(t)$ into a bunch of terms, one of which is (some scalar multiple of) $e^{2 \pi i f t}$.

Now the Fourier transform of this signal is $$ \int_0^1 e^{2 \pi i f t} e^{-2 \pi i f t} \, d t = \int_0^1 1 \, d t = 1. $$

On the other hand, if we try this with some frequency not present in the signal, say $f' \neq f$, we get $$ \int_0^1 e^{2 \pi i f' t} e^{-2 \pi i f t} \, d t = \int_0^1 e^{2 \pi i (f' - f) t} \, d t = \frac{e^{2 \pi i (f' - f) t}}{2 \pi i (f' - f)} \biggr\rvert_0^1 = 0. $$

Note that the integral in the opening post (and in the video) is missing the bounds of integration, which in this case are very important.