= Solution
Cut the octagon for $S_3$ along a diagonal joining the appropriate two vertex classes. The two resulting polygons can be reattached along that diagonal in the opposite order. Reading the new boundary word gives the square word
$$
e f^{-1} e f,
$$
which is exactly the edge identification displayed for $S_4$. The cut-and-paste map is affine on the two pieces and respects every paired edge, so it descends to a <homeomorphism> $S_3\cong S_4$. Both are the <Klein bottle>.
After deleting an open disc, the punctured Klein bottle can be realized as a boundary connected sum of two embedded <Möbius band>[Möbius strips], and hence embeds in $\mathbb R^3$. The closed Klein bottle cannot embed: every connected closed surface embedded in $\mathbb R^3$ is two-sided and therefore orientable, by the <Jordan-Brouwer separation theorem>, whereas the Klein bottle is nonorientable. Thus
$$
\boxed{S_4\setminus D^2\text{ embeds in }\mathbb R^3,
\qquad S_4\text{ does not}.}
$$
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