Is there a partially order relation on $S^1$ whose corresponding order topology coincides to the standard topology of circle?Is there a totally order relation with such property?
what is a theory devoted to such questions for compact topological groups?
Yes, there's such a poset (not a totally ordered one, as already observed).
Here, if $X$ is a poset I understand that the interval topology is the topology defined by a prebasis consisting of those $U_{x_0}=\{x:x>x_0\}$ and $V_{y_0}=\{y:y<y_0\}$ where $x_0,y_0$ ranges over $X$.
Now define a set as follows: $$X=\{-\infty\}\sqcup(\mathbf{R}\times\{-1,1\})\sqcup\{+\infty\}$$
It has a natural topology making it homeomorphic to the circle ($(x_i,n_i)$ tends to $\pm\infty$ meaning that $x_i\to\pm\infty$ in the usual sense).
Now make it a poset as follow: $-\infty\le z\le +\infty$ for all $z$; on each of the two copies of $\mathbf{R}$, we take the usual ordering. Finally, for $e\in\{1,-1\}$, we say ($\sharp$) that $(x,e)\le (y,-e)$ if $x\le y-1$.
Then we see that the interval topology coincides with the previous one. At $\pm\infty$, this is clear: for each finite set $F$ with $-\infty\notin F$, there exists a neighborhood of $-\infty$ that is $\le y$ for each $y\in F$. At $(x,e)$, observe that for $t<1$, $\{z: (x-t,e)\le z\le (x+t,e)\}$ is reduced to $\{(y,e):x-t\le y\le x+t\}$. (Note that this uses the $-1$ in ($\sharp$)).