"Let $K$ be (the topological space that is known to topologists as) the Klein bottle. There's a standard immersion $f:K \to \mathbb{R}^3$, whose image is known, in popular culture, as "the Klein bottle". $f$ is, of course, not an embedding. Is there a continuous injective function defined on a product of intervals, $\sigma:I \times J \to \mathbb{R}^3$, such that $\sigma(I \times J) = f(K)$?''
I think no, because Klein bottle intersects itself. But I think it is not a safe justification, since the cillinder
\begin{cases} 25x^2 - 25y^2 - x^4 - 2x^2y^2 - y^4 = 0 \\ -1 \leq z \leq 1 \end{cases}
(generated by traslating a lemniscate along the $z$ axis) in some way intersects itself, but it is the image of a continuous, injective $\sigma: (-1,1) \times [-1,1] \to \mathbb{R}^3$, as indicated in the figure below. I still think my conclusion about the Klein bottle is true, but I wish a better justification for it.
Yes, it's possible, inspired by your "extruded figure 8" example.
Look at the place where the "tube" crosses the fat surface; the two intersect in a circle, which bounds a disk in the "fat surface." This is the brownish-orange circle in the top-left portion of the figure below.
Remove this circle from the "tube" part, but leave it in the "fat surface" (just as in the extruded figure-8 example, you removed the center-line from one sheet but not the other). This results in the second figure: the brown circle remains as part of the main body; the "tube ends" are both "open" (i.e., they look like $S^1 \times (0, ...]$),and I've drawn them dotted. In the next three stages, I've just done some nice homotopies to make this look (in the fourth picture) like $S^1 \times (-1, 1)$. In the fifth picture, I've cut apart the along one of the generating lines, so that in the last picture we have $(-\pi, \pi) \times (-1, 1]$, which is a product of intervals, and we're done.