I don't mean to be a bother.
It seems as though the answer may be obvious, but then, seemingly simple math questions can have surprising answers.
I should also like any pointers re: the general case for the torus.
Thanks in advance for any (polite) suggestions.
For any $d\geq1$ a $d$-dimensional torus $T^d$ is a $d$-dimensional manifold whose most basic model is ${\mathbb T}^d:={\mathbb R}^d/{\mathbb Z}^d$. Think of a $d$-dimensional cube whose opposite $(d-1)$-dimensional faces are identified in pairs. The one-dimensional torus ${\mathbb T}={\mathbb R}/{\mathbb Z}$ happens to be the same thing as the one-sphere, or circle, $S^1$. The two-dimensional torus appears in many computer games as a rectangular board where creatures disappearing on the right surprisingly reappear on the left moving in the same direction.
A familiar realization of a $2$-torus $T^2$ is the surface $$S:\quad \bigl(\sqrt{x^2+y^2}- a\bigr)^2 + z^2=b^2,\qquad a>b>0,$$ in ${\mathbb R}^3$. A parametric representation of $S$ is given by $$(\phi,\theta)\mapsto\bigl((a+b\cos\theta)\cos\phi, \>(a+b\cos\theta)\sin\phi, \>b\sin\theta\bigr)\qquad\bigl(\phi\in{\mathbb R}/(2\pi{\mathbb Z}), \ \theta\in{\mathbb R}/(2\pi{\mathbb Z})\bigr)\ ,$$ which exhibits $S$ as bijective copy of the basic model ${\mathbb T}^2$ of such a surface.
The donut you can eat, i.e., the $3$d body $$B:=\bigl\{(x,y,z)\>\bigm|\> \bigl(\sqrt{x^2+y^2}- a\bigr)^2 + z^2\leq b^2\bigr\},\qquad a>b>0,$$ is called a full torus.
The three-dimensional object you have in mind is not the three-dimensional torus $T^3\sim T^1\times T^1\times T^1$, but is the product of a $2$-sphere $S^2$ with $T^1$. You can envisage this object in the form of a spherical shell (with positive thickness) in ${\mathbb R}^3$ whose inner and outer spherical surfaces have been identified in the obvious way. Going around the $x_4$-axis in your hypothetical picture then corresponds to traversing the shell outwards in radial direction and automatically being teleported to the inner boundary when arriving at the outer boundary.
I'm way late to the party as usual, but I've got an equation for you, and some nice visuals of this object. Christian already gave the specifics a few ways. As you know, a torus can be constructed by starting with a circle in the xy plane:
$$x^2 + y^2 = r^2$$
Translate the circle by R units along the x-axis, which should be around $R = 2\cdot r$, for a ring torus,
$$(x-R)^2 + y^2 = r^2$$
Then, sweep the circle around a circular path into 3D, by replacing $x$ with $\sqrt{x^2+z^2}$ ,
$$\left(\sqrt{x^2+z^2}-R\right)^2 + y^2 = r^2$$
which gives us the equation of a torus.
To construct the $S^2$ x $S^1$ , we start with a sphere in plane $xyz$:
$$x^2 + y^2 + z^2 = r^2$$
Shift along $x$ by $R$ units where $R = 2\cdot r$,
$$(x-R)^2 + y^2 + z^2 = r^2$$
Sweep around in a circle, into 4D:
$$\left(\sqrt{x^2+w^2}-R\right)^2 + y^2 + z^2 = r^2$$
Cross Sections of $S^2$ x $S^1$
Taking some sections of this object, we can set either $w=0$ , or $z=0$:
$$\left(\sqrt{x^2}-R\right)^2 + y^2 + z^2 = r^2$$
This cross-section will give us the disjoint pair of spheres that you're looking for, centered at $(+R,0,0)$ and $(-R,0,0)$, since $\left(\sqrt{x^2}-R\right)^2$ has two values of $(x-R)^2$ and $(x+R)^2$
$$\left(\sqrt{x^2+w^2}-R\right)^2 + y^2 = r^2$$
This cross-section will give us a single torus.
Now, what does this thing look like, as sliced in 3D? Using the rotate equation,
$$\left(\sqrt{x^2+\left(w\cdot \cos(t)+z\cdot \sin(t)\right)^2}-R\right)^2 + y^2 + \left(w\cdot \sin(t)-z\cdot \cos(t)\right)^2 = r^2$$
Animating the value of t from 0 to 2$\pi$:
Holding at different angles, and passing through a 3-plane:
