The Fourier series is a way of building up functions on $[-\pi,\pi]$ in terms of functions that diagonalize differentiation--namely $e^{inx}$. If $L=\frac{1}{i}\frac{d}{dx}$ then $Le^{inx}=ne^{inx}$. That is $e^{inx}$ is an eigenfunction of $L$ with eigenvalue $n$. The fact that all square integrable functions on $[-\pi,\pi]$ can be expanded as $f = \sum_{n=-\infty}^{\infty}c_n e^{inx}$ is quite a nice thing. If you want to apply the derivative operator $L$ to $f$, you just get $Lf = \sum_{n=-\infty}^{\infty}nc_ne^{inx}$. More generally, if $f$ has $N$ square integrable derivatives, then the $N$-th derivative is
$$\frac{1}{i^{n}}f^{(n)}=L^{N} f = \sum_{n=-\infty}^{\infty}n^{N}c_n e^{inx}.$$
Diagonalizing an operator makes it easier to solve all kinds of equations involving that operator. The only issue is this: How do you find the correct coeffficients $c_n$ so that you can expand a function $f$ in this way? For the ordinary Fourier series,
$$
c_n = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(t)e^{-int}dt.
$$
On a finite interval, this is great. But what happens if you want to work on the entire real line? If you work on larger and larger intervals, then you get more and more terms. You need terms with larger and larger periods, and all multiples of those. In the limit of larger intervals, you need an integral to sum up all the terms, with every possible periodicity. That is, you can expand a square integrable $f$ as
$$f(x) = \int_{-\infty}^{\infty}c(s)e^{isx}ds.
$$
As before, applying operations of $L=\frac{1}{i}\frac{d}{dx}$ is easier using this representation of $f$:
$$
\frac{1}{i^{N}}f^{(N)}(x)= L^{N}f = \int_{-\infty}^{\infty}s^{N}c(s)e^{isx}ds.
$$
You can see that the discrete and the continuous cases are remarkably similar. And, based on that, how might you expect to be able to find the coefficient function $c(s)$? As you might guess,
$$
c(s) = \frac{1}{2\pi}\int_{-\infty}^{\infty}f(x)e^{-isx}dx.
$$
The Fourier transform is a way to diagonalize the differentiation operator on $\mathbb{R}$.
The reason that the discrete and continuous Fourier transforms are so important is that they diagonalize the differentiation operator. One way to view the effects of diagonalization is that you turn the operator into a multiplication operator. You can see how that makes solving differential equations a lot easier. In the coefficient space all you do is to divide in order to invert.
It's the same way with a matrix: if you have a big matrix equation
$$Ax = y,$$
and if $A$ is symmetric, then you can find a basis $\{ e_1,e_2,\cdots,e_n \}$ where $Ae_k = \lambda_k e_k$. Then, if you can expand $x$ and $y$ in this basis
$$x = \sum_{k=1}^{n} c_k e_k \\y = \sum_{k=1}^{n} d_k e_k.$$
Then the new equation is solved by division:
$$Ax = y \\ \sum_{k=1}^{n} c_k \lambda_k e_k = \sum_{k=1}^{n} d_k e_k \\
c_k = \frac{1}{\lambda_k} d_k e_k.$$
So if you know how to expand $y$ in the $e_k$ terms as $\sum_{k}d_k e_k$, then you can get the solution $x$ by division on the coefficients
$$x = \sum_{k=1}^{n} \frac{1}{\lambda_k} d_k e_k$$
(Assuming none of the $\lambda_k$ are $0$.)
The discrete and continuous Fourier transforms are a way to diagonalize differentiation in an infinite-dimensional space. And that allows you to solve linear problems involving differentiation.
Hilbert Transform: The Hilbert transform was developed by Hilbert to study the operation of finding the harmonic conjugate of a function. For example, the function $f(z) = z^2=(x+iy)^2=x^2-y^2+i(2xy)$ has harmonic real and imaginary parts. Hilbert was trying to find a way to go between these two components (in this case $x^2-y^2$ to $2xy$.) The setting of this transform is the upper half plane. If you start with a function $f(x)$, find the function $\tilde{f}(x,y)$ that is harmonic in the upper half plane, and then find $g(x,y)$ such that $f(x,y)+ig(x,y)$ is holomorphic in the upper half plane, then the Hilbert transform maps $f$ to $g$.
Because $i(f+ig)=-g+if$ is also holomorphic, then the transform maps $g$ to $-f$, which means that the square of the transform is $-I$. In this setting, the Hilbert transform turns out to be concisely expressed in terms of the Fourier transform if you work with square integrable functions.
