I will call the $m$-covered circle the map $f:[0,2\pi]\to \mathbb{R}^2$ defined by $f(\theta)=(\cos m\theta,\sin m\theta)$. This is an immersion of the unit circle in $\mathbb{R}^2$. The image of $f$ is the the same as the image of the standard embedding of the unit circle in $\mathbb{R}^2$. However, the difference is that $f$ has degree $m$ as a map $S^1\to S^1$. (Note that $f$ it is not an embedding of $S^1$ in $\mathbb{R}^2$ since $f$ is not injective).
How can I parameterize an $m$-covered $n$-sphere? In other words, I am looking for an immersion $F_m:S^n\to \mathbb{R}^{n+1}$ whose image is the unit $n$-sphere in $\mathbb{R}^{n+1}$ and such that $\mathrm{deg}(F_m)=m$.
My attempt for $n=2$: The standard embedding $[0,2\pi]\times[0,\pi]\to\mathbb{R}^3$ is given by $$(\theta,\phi)\mapsto (\sin\phi\cos\theta,\sin\phi\sin\theta,\cos\phi).$$ Now define $F_m:[0,2\pi]\times[0,\pi]\to \mathbb{R}^3$ by $$F_m(\theta,\phi)\mapsto (\sin\phi\cos m\theta,\sin\phi\sin m\theta,\cos\phi).$$ Does this work? Thanks for your help.
You can take the $n-1$-fold suspension of your map. To do so, you'll need to read about suspensions, of course, but that really is the whole story.
For instance, we can write $s_m : S^1 \to S^1$ for the map you called $f$. Writing
$$ S^2 = S^1 \times [-1, 1] / \sim, $$ where $(P, 1) \sim (Q, 1)$ for any $P, Q \in S^1$, and
where $(P, -1) \sim (Q, -1)$ for any $P, Q \in S^1$. Then the map $s_2$ using coordinates on $S^1 \times [-1, 1]$, is just (the quotient of) $$ (P, t) \mapsto (s_1(P), t) $$ Now you can iterate this process to get a map from $S^3 \to S^3$, using essentially exactly the same formula, but replacing $s_1$ with $s_2$, and so on.