Basic question about trigonometric series and transforms thereof
There is no general relationship between the Fourier transform of $f$ and that of $g(f)$ where $g$ is an arbitrary function. The Fourier transform does have the linearity property, so if $g$ is something simple like an affine transform, then the same linear relationship applies to their transforms $F$ and $G$.

With respect to your second question, where $h = \sum_k \left(A_k \sin(\omega_k x)\right)^N$, the presence of more than $k$ peaks in the Fourier transform of $h$ is easily explained using the power-reduction trigonometric identity:

$$
\sin^n(\theta) = 
\begin{cases} 
\frac{2}{2^n} \sum_{k=0}^{\frac{n-1}{2}} (-1)^{(\frac{n-1}{2}-k)} \binom{n}{k} \sin{((n-2k)\theta)}, &amp; n \text{ is odd} \\
\frac{1}{2^n} \binom{n}{\frac{n}{2}} + \frac{2}{2^n} \sum_{k=0}^{\frac{n}{2}-1} (-1)^{(\frac{n}{2}-k)} \binom{n}{k} \cos{((n-2k)\theta)}, &amp; n \text{ is even}
\end{cases}
$$

(the above is shamelessly borrowed from Wikipedia)

So, when you raise a sinusoid to a power, the result can be expressed as a weighted sum of sinusoids at different frequencies, where the number of individual terms is related to the power. That's why you see additional peaks in the spectrum of $h$.

You can come up with a more general relationship for some cases by taking advantage of the multiplication property of the Fourier transform. That is, if $g = f \cdot e$, then its Fourier transform is $G = F * E$ (where $*$ indicates convolution). You could apply this relationship repeatedly to the sinusoid raised to a power to derive the same result as above.