(This answer is told mainly from the group action point of view, because it's what I'm most comfortable with. I'm planning on posting a different answer with a more geometric feel to it shortly.)
There are two "main" foliations of $S^3$ and I think you're confusing the two.
The first comes from the action of $S^1$ on $S^3$. To describe this action, think of $S^1$ as the unit complex numbers and $S^3 = \{(z_1,z_2)\mid |z_1|^2+|z_2|^2 = 1\}$. Then for $z\in S^1$, we have $z\ast(z_1,z_2) = (zz_1,zz_2)$. This action is the (relatively) famous Hopf action. Through every point, the orbit is a circle.
To see this, note that the orbit is always diffeomorphic to a quotient of $S^1$ by a closed subgroup of $S^1$, an hence is diffeomorphic to $S^1$ or a point. The only way the orbit through $(z_1,z_2)$ could be a point is if $zz_1 = z_1$ and $zz_2 = z_2$ for every unit complex number $z$. This is equivalent to $z_1(z-1) = 0$ and $z_2(z-1) = 0$ for all $z$. This implies that both $z_1$ and $z_2$ are $0$, contradicting the fact that $(z_1,z_2)\in S^3$.
The second foliation comes from the action of $S^1\times S^1$ on $S^3$. To describe this action, we think of $S^1\times S^1$ as the collection of pairs of unit complex numbers. Then $(z,w)\in S^1\times S^1$ acts on $S^3$ by $(z,w)\ast(z_1,z_2) = (zz_1,wz_2)$. As before, the orbit through a point is always diffeomorphic to $S^1\times S^1$ quotiented out by a closed subgroup, so is diffeomorphic to either $S^1\times S^1$, $S^1$, or a point.
If "generic" means "both $z_1$ and $z_2$ are nonzero", then the orbit through any generic point is $S^1\times S^1$. To this note that no nontrivial element of $S^1\times S^1$ fixes any generic point, for is $(z,w)\ast(z_1,z_2) = (z_1,z_2)$, then we must have $z_1(z-1) = z_2(w-1) = 0$. Since $z_1\neq 0 \neq z_2$, we must have $z=w=1$, i.e., $(z,w)$ is trivial.
For nongeneric points (either $z_1 = 0$ or $z_2 = 0$), the orbit is a circle. For, suppose $z_1 = 0$. Then we have $(z,1)\ast(0,z_2) = (0,z_2)$, meaning all points in $S^1\times\{1\}$ fix $(0,z_2)$. In fact, these are all the points that fix $(0,z_2)$. It follows that the orbit is diffeomorphic to $S^1\times S^1/ S^1\times \{1\} \cong S^1$.
Similarly, if $z_2 = 0$, the orbit is diffeomorphic to $S^1\times S^1 / \{1\}\times S^1\cong S^1$.
Putting this altogether, the second action gives a foliation on $S^3$ deomposing it as a disjoint union of tori, except for two exceptional circles.