I've seen several derivations of the Fourier transform, but most don't cover the conversion to the form
$$ S(f) = \int_{\infty}^{-\infty} s(t)e^{-i2\pi ft} \;\mathrm{d}t $$
What is the role of each of the parts of the exponent to $e$ ($i, 2\pi, f, $ and $ t$) in breaking up one composite wave into component sine waves? $2\pi$ is most likely something to do with circles I take it, but what is its specific role in mapping one to the other, and how is its presence seen in the result of Fourier transform in producing the component sine waves?
An ideal answer also step that introduces -i to the expression.
The $2 \pi$ is not essential; that just converts between angular frequency and ordinary frequency. There are definitions of the Fourier transform which don't use this factor at all. These are common in PDE theory.
The gist of what's going on is very much like the situation in Fourier series: if you have a complex-valued square integrable function $f$ and a complex-valued square integrable function $g$, then the orthogonal projection of $f$ onto the span of $g$ is $g \frac{\int_{-\infty}^\infty f(x) \overline{g(x)} dx}{\int_{-\infty}^\infty |g(x)|^2 dx}$.
If $F$ is the Fourier transform of $f$ then $F(\xi)$ is "morally" the coefficient in the projection of $f$ onto $e^{i 2 \pi \xi t}$. (The minus sign comes from the conjugate above, since in general $\overline{e^{ix}}=e^{-ix}$.) But note that we do not divide by $\int_{-\infty}^\infty |e^{i 2 \pi \xi t}|^2 dt$. This is because this is actually $+\infty$, which would be a problem. But this is the idea. The inversion formula makes this idea precise.
You might be asking "why do we want to project onto the span of $e^{i 2 \pi \xi t}$?" One answer is that, because of Euler's identity, it is a compact way of projecting onto sinusoids, which plays nicer with derivatives. Sinusoids are what we want to use to think about frequency, which is what the Fourier transform is all about.