There is a theorem that states that if $f$ is analytic in a domain $D$, and the closed disc {$  z:|z-\alpha|\leq r$} contained in $D$, and $C$ denotes the disc's boundary followed in the positive direction, then for every $z$ in the disc we can write:
$$f(z)=\frac{1}{2\pi i}\int\frac{f(\zeta)}{\zeta-z}d\zeta$$

My question is:
What is the intuitive explanation of this formula? (For example, but not necessary, geometrically.)

(Just to clarify - I know the proof of this theorem, I'm just trying to understand where does this exact formula come from.)

If you are looking for intuition then let us assume that we can expand $f(\zeta)$ into a power series around $z$: $f(\zeta) = \sum_{n \geq 0} c_n(\zeta - z)^n$. Note $c_0 = f(z)$.
If you plug this into the integral and interchange the order of integration
and summation then that integral on the right side of the formula
becomes $\sum_{n \geq 0} \int c_n(\zeta - z)^{n-1}d\zeta$.  Let us also assume that an integral along a contour doesn't change if we deform the contour continuously through a region where the function is "nice". So let us take as our path of integration a circle going once around the point $z$ (counterclockwise). Then you are basically reduced to
showing that $\int (\zeta - z)^{m}d\zeta$ is 0 for $m \geq 0$ and is $2\pi i$ for $m = -1$.
These can be done by direct calculations using polar coordinates with $\zeta = z + e^{it}$.  Now divide by $2\pi i$ and you have the formula.  Of course this is a hand-wavy argument in places, but the question was not asking for a rigorous proof.  Personally, this is how I first came to terms with understanding how Cauchy's integral formula could be guessed.