Here's a description of an embedding, which can be elaborated into an
explicit formula if you like.
Take an embedding of $T^n$ into $\Bbb R^{n+1}$. We want to use this
as a basis for one of $T^{n+1}=T^n\times S^1$ into $\Bbb R^{n+2}$.
Translate your $T^n$ so that it lies within the half-plane
defined by $x_{n+1}>0$. Then let the embedding of $T^n$
be given by functions
$p\mapsto (\phi(p),\psi(p))$ where $\phi(p)\in\Bbb R^n$ and $\psi(p)\in(0,\infty)$. Now the embedding of $T^{n+1}$ into $\Bbb R^{n+2}$ by
$$(p,e^{it})\mapsto(\phi(p),\psi(p)\cos t,\psi(p)\sin t).$$
As $\psi(p)>0$ the last two coordinates determine $\psi(p)$ and $e^{it}$.
Had some enforced free time today, so I gave some more thought to this. I know I've seen a much nicer argument, but it has been too long. This is what I came up with:
Suppose we have such a smooth chart $f: \Bbb R^2 \supset U \to S^2$. Find a simply-connected open set $D \subset U$, maximal with respect to $f(D)$ being 1-1. Any such set $D$ is diffeomorphic to the open unit disk, so it is sufficient to assume that $D$ actually is the unit disk. We can extend $f$ to $\overline D$, and necessarily $f(\overline D) = S^2$. $f$ is injective on the interior, but on the unit circle every point must have at least one other point with the same image under $f$ (otherwise $D$ would not be maximal).
Let $p \in f(S^1)$. If $f^{-1}(p)$ is dense in any arc of $S^1$, then by continuity $f$ must be constant on the arc, which would violate that $f$ is a chart. So $f^{-1}(p)$ is nowhere dense. Choose a maximal open arc $\overset{\frown}{AB}$ of $S^1$ with $p \notin f(\overset{\frown}{AB})$ (so $f(A) = f(B) = p$). The image of the chord $\overline {AB}$ under $f$ is a simple closed loop in $S^2$ passing through $p$. As such, it divides the sphere into two separate regions. The portion of the disk on the arc-side of $\overline {AB}$ is mapped to the interior of that loop, while the other side is mapped to the exterior. By continuity, $\overset{\frown}{AB}$ must map to the interior of the loop.
Let $C \in \overset{\frown}{AB}$, and $C' \ne C$ be any other point of $S^1$ with $f(C') = f(C)$. If $C' \notin \overset{\frown}{AB}$, then a line segment connecting a point between $\overline{AB}$ and $\overset{\frown}{AB}$ with $C$, combined with a line segment connecting $C'$ with a point in the open disk on the other side of $\overline{AB}$, is mapped by $f$ to a continuous curve connecting a point interior to the loop to a point exterior to the loop without every crossing the loop, which cannot be. Therefore for any $C \in \overset{\frown}{AB}$, it must be that $f^{-1}(f(C)) \subset \overset{\frown}{AB}$.
Since every point of $f(S^1)$ has multiple pre-images, we can take $A'$ to be the midpoint of $\overset{\frown}{AB}$ and $B\,'$ any other point of $\overset{\frown}{AB}$ with $f(B\,') = f(A')$, and follow the same argument again. Continuing this way we get a decreasing sequence of arcs that must converge to some point $C$ which has to be a singularity of $f$, since moving from $C$ along the unit circle in either direction will under $f$ move in the exactly the same direction.
Therefore no chart $f: \Bbb R^2 \supset U \to S^2$ can be surjective.
I've skimmed over a lot of details, and it is certainly possible that one of them may prove more of a problem than I realize, but I think this works.
Best Answer
Start with the equation $$x_1^2+x_2^2=r_1^2$$ which describes a centered circle in the place of radius $r_1$. Replace $x_2$ by $\sqrt{x_2^2+x_3^2}-r_2$, so get $$x_1^2+(\sqrt{x_2^2+x_3^2}-r_2)^2=r_1^2.$$ This is a $2$-torus with radii $r_1$ and $r_2$. Let's do this again: replace the last variable $x_3$ by $\sqrt{x_3^2+x_4^2}-r_3$, to get $$x_1^2+(\sqrt{x_2^2+(\sqrt{x_3^2+x_4^2}-r_3)^2}-r_2)^2=r_1^2.$$ This is a $3$-torus with radii $r_1$, $r_2$, $r_3$.
One can go on as long as one wants.
This is just using the procedure which goes from, say, a curve to the surface of revolution it generates, several times.
(Of course, the radii have to be such that the thing is an actual torus, without any self-intersections)