I'm a bit in truble with the circle.
Question 1
I want to use $\mathcal C=\{e^{it}\mid t\in [0,2\pi]\}$. Since it's the image of $[0,2\pi]$ under the coninuous function $t\mapsto e^{it}$, the set $\mathcal C$ is compact. But is it a metric space ? I know that I can put the euclienne metric, but I would like to use the metric $$d(e^{it},e^{is})=|t-s|.$$
I proved it was a metric but $t,s\in [0,2\pi)$ only because otherwise, we have $d(e^0,e^{2i\pi})=|2\pi|>0$. So, to use this metric, I have to consider $\mathcal C=\{e^{it}\mid t\in[0,2\pi)\}$, no ? And with this definition, it's not compact anymore, right ?
Question 2
Now if I want to stud periodic function, I know that any $2\pi-$periodic function can be seen as a function on the unit circle, i.e.
$f:\mathbb R\to \mathbb R$ is a $2\pi-$periodic function $\iff$ there is a function $F:\mathcal C\to \mathbb R$ s.t. $f(\theta )=F(e^{i\theta })$.
So, for example, $f$ is continuous $\iff$ $F$ is continuous or $f$ is derivable $\iff$ $F$ is derivable. But which topology I have to use on $\mathcal C$ to talk about continuity ? And for the derivability, I don't really understand how can such a function $F$ be derivable on the Circle... What would be the sense of $F'$ ?
Question 1: It is allowed to produce the set ${\cal C}$ as bijective image of the map $$\psi:\quad [0,2\pi[\>\to{\mathbb C},\qquad t\mapsto e^{it}\ ,$$ and then to define a topology on ${\cal C}$ via your metric $$d(e^{it},e^{is}):=|t-s|\ .$$ But this is not the topology we want on ${\cal C}\,$! The proper reason we introduced the map $t\mapsto e^{it}$ here is that the euclidean topology on ${\cal C}$ exactly reflects what our intuition wants, namely that ${\cal C}$ should be a closed loop, and not a half-open interval. Now the euclidean distance $$d_{\rm eucl}(e^{it},e^{is}):=|e^{it}-e^{is}|$$ is not always agreeable in computations. You can always replace it by the "equivalent" $$\hat d(e^{it},e^{is}):=\min_{k\in{\mathbb Z}}|t-s+2k\pi|\ .$$ This $\hat d$ gives the length of the shorter arc on ${\cal C}$ connecting $e^{it}$ with $e^{is}$, whatever $s, t\in{\mathbb R}$.
Question 2. Your outlined statement is correct. – If $f$ is a $2\pi$-periodic function on ${\mathbb R}$ then the function $$F:\quad {\cal C}\to{\mathbb R},\qquad z\mapsto F(z):=f\bigl({\rm arg}(z)\bigr)\tag{1}$$ is well defined on the curve ${\cal C}$. Here ${\rm arg}(z)$ is the polar angle of $z$ up to integer multiples of $2\pi$. This multivaluedness of ${\rm arg}$ is of no harm here, since $f$ is $2\pi$-periodic. Instead of $(1)$ one can use the formula $F(e^{i\theta})=f(\theta)$ as "definition" of $F$. Again the periodicity of $f$ guarantees that $e^{i\theta}=e^{i\theta'}$ when $\theta-\theta'=2k\pi$ causes no difficulties.
Furthermore it is true that the function $F:\>{\cal C}\to{\mathbb R}$ is continuous on the set ${\cal C}\subset{\mathbb C}$ (with the euclidean topology) iff the $2\pi$-periodic function $f$ is continuous on ${\mathbb R}$. Continuity of $F$ means that $$\lim_{z\to z_0} F(z)=F(z_0)\qquad(z, \>z_0\in{\cal C})\ .$$
But you should not think about the differentiability of $F$ as a function on the $(x,y)$-space ${\mathbb R}^2\sim{\mathbb C}$. The function $F$ is, however, differentiable on the one-dimensional manifold ${\cal C}$ iff $f$is differentiable on ${\mathbb R}$, because of the local coordinate $\theta$ established on ${\cal C}$ by $\theta\mapsto x+iy:=e^{i\theta}$.