This question belongs to a series of questions that keep arising in my mind from time to time: here and here are (in order) the two questions that I asked before.
The aim of the first question, let's say finding a similar "parametric definition" that was conformal for tori in $\mathbb{S}^3$ as that one in $\mathbb{R}^3$, was apparently too optimistic. Not even what I thought was a "parametric definition" in $\mathbb{R}^3$ seemed to be correct in the answer I got. Then I asked in the second question if it was possible to start with a closed curve and by rotating it construct a torus in $\mathbb{S}^3\subset\mathbb{R}^4$. I didn't get answers to this question.
At this point I'd like just to understand the most basic ideas. The parametric definition in $\mathbb{R}^3$ of a torus comes from a construction using a circle and rotating it. Suppose I start with a circle $$(x-a)^2+z^2=r^2,$$ for some fixed $a>r>0$.
Parametrizing this curve and multipliying the parametric vector by a matrix like $$\begin{pmatrix} \cos\phi & -\sin\phi & 0\\ \sin\phi & \cos\phi & 0\\ 0 & 0 & 1\end{pmatrix}$$ we rotate the curve around the $z$-axis (visually is easy to see that we construct in this way a torus in $\mathbb{R}^3$) and the usual parametric definition that one can find in the literature is obtained.
Now I'd like to imitate this idea but in $\mathbb{R}^4$. So I start with the same circle $(x-a)^2+z^2=r^2$, I take a parametrization $$x=r\cos\theta+a$$ $$z=r\sin\theta,$$ where $r\in(0,a)$ and $\theta\in[0,2\pi]$. Now I rotate it by doing $$\begin{pmatrix} \cos\phi & -\sin\phi & 0 & 0\\ \sin\phi & \cos\phi & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\end{pmatrix}\cdot\begin{pmatrix}r\cos\theta+a\\ 0\\ r\sin\theta\\ 0\end{pmatrix}=\begin{pmatrix}\cos\phi(r\cos\theta+a)\\ \sin\phi(r\cos\theta+a)\\ r\sin\theta\\ 0\end{pmatrix}$$
This construction is exactly the same as that one in $\mathbb{R}^3$, which makes sense completely. Because of this I see that my simple question should be:
What is a torus (not topologically, that is clear) in $\mathbb{R}^4$? And once that's clear, what would be the analogous process to 'fit' a torus in $\mathbb{R}^4$? From there should be easy to understand how would it be in $\mathbb{S}^3\subset\mathbb{R}^4$.
This construction can never produce a torus in $\mathbb{S}^{3}_{r}$ for any $r>0$. Since rotations with centre $0$ act on $\mathbb{S}^{3}_{r}$, the torus you construct will be contained in $\mathbb{S}^{3}_{r}$ if and only if the original circle is. The circle contains the points $(a-r,0,0,0)$ and $(a+r,0,0,0)$, which have different distance from the origin, so this is not the case.
(Notation: $\mathbb{S^{3}}_{r} = \{v \in \mathbb{R}^{4} : v \cdot v = r^2\}$)
Edit: case $a = 0$. in this case the point $(0,0,r,0)$ is fixed by the rotation, so the space you obtain is not homeomorphic to torus. In fact the space you obtain is homeomorphic to a $2$-sphere.
The standard way to embed a torus in $\mathbb{S}^{3} \subset \mathbb{C}^{2}$ is by the equaction $|z_{1}|^2 = |z_{2}|^{2} = \frac{1}{2}$ this is the so called Clifford torus. https://en.wikipedia.org/wiki/Clifford_torus.