There are three ways to think about a function on the unit circle, as explained in the first video:

1. It's a function on a geometric set, and we can give it the inner product $\langle {\bf f} | {\bf g} \rangle =\int_{S^1} \bar f g ds$, where $s$ is arclength. I have intentionally avoided writing a variable for $f$ and $g$, because there are many ways to parametrize the circle.
2. We can parametrize our circle by $x = \cos(\theta)$, $y=\sin(\theta)$, with $\theta$ running from $-\infty$ to $\infty$. Of course, $\theta$ and $\theta + 2\pi$ refer to the same point, so we have a periodic function $f$ on $\mathbb{R}$ with $f(\theta+2\pi)=f(\theta)$. Our inner product comes from integrating over just one period: $\langle {\bf f} | {\bf g} \rangle =\int_0^{2\pi} \overline{f(\theta)} g(\theta) d\theta$.
3. Finally, we can think of a function on the interval $[0,2\pi]$, but insist that the values and derivatives match at the endpoints: $f(0)=f(2\pi)$, $f'(0)=f'(2\pi)$, etc.

Now let's switch to functions with period $L$ instead of period $2\pi$. Functions in the space $L^2(S^1_L)$ can either be viewed as

1. Functions on a circle of circumference $L$.
2. Periodic functions on $\mathbb{R}$ with $f(x+L)=f(x)$, or
3. Functions on $[0,L]$ with periodic boundary conditions $f(0)=f(L)$, $f'(0)=f'(L)$, etc. [Strictly speaking we can't assume that our functions are differentiable, or even continuous. But if the one-sided derivatives at $0$ and $L$ disagree, then this is like a periodic function where the derivative takes a jump at $x=0$.]

Now that we have our function space, we can look at some operators. $A = d^2/dx^2$ and $P = -id/dx$ are both Hermitian, so we can find an orthogonal basis of eigenvectors. The eigenvalues of $A$ are $0$ (with multiplicity 1) and $-4n^2\pi^2/L^2$ (with multiplicity 2). The eigenvectors are $\cos(2\pi n x/L)$ with $n=0,1,2,\ldots$ and $\sin(2 \pi n x/L)$ with $n=1, 2,\ldots$. This is explained in the following video:

The eigenvalues of $P$ are $2 \pi n/L$, where now $n$ can be positive or negative or zero. The eigenvectors are $\exp(2 \pi i n x/L) = \cos(2 \pi n x/L ) + i \sin(2 \pi n x/L)$. Diagonalizing $P$, and the vectors we get that way, are explained in the third video.

From these orthonormal bases we get two more ways to decompose a function. If $f(x)$ is a periodic function with period $L$, then we can write $$f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty \left [ a_n \cos\left ( \frac{2 \pi n x}{L}\right ) + b_n \sin\left ( \frac{2 \pi n x}{L}\right )\right ],$$ where \begin{eqnarray*} a_n & = & \frac{2}{L}\int_0^L f(x) \cos \left ( \frac{2 \pi n x}{L}\right )dx , \cr b_n & = & \frac{2}{L}\int_0^L f(x) \sin \left ( \frac{2 \pi n x}{L}\right )dx. \end{eqnarray*} We can also write $$f(x) = \sum_{n=-\infty}^\infty c_n \exp(2 \pi i n x/L), \qquad \hbox{with}$$ $$c_n = \frac{1}{L} \int_0^L f(x) \exp(-2\pi i n x/L) dx.$$ The coefficients $c_n$ are often written $\hat f_n$ and are called the Fourier coefficients of $f$.