$S_4/V_4$ isomorphic to $S_3$ - Understanding Attached Tables

I'm guessing that by $\,Z_4\,$ you actually mean the Klein viergrupp

$$C_2\times C_2\cong \{(1)\,,\,(12)(34)\,,\,(13)(24)\,,\,(14)(23)\}$$

since the above one is the only normal subgroup of order $\,4\,$ of $\,S_4\,$ , but then, since the cyclic group of order $\,6\,$ is obviously abelian ,we get

$$S_4/Z_4\cong C_6\Longleftrightarrow S_4^{'}\leq Z_4$$

and this is false since $\,(123)\notin Z_4\,$ , but $\,(123)\in A_4=S_4^{'}\,$.

Thus, as the only other group of order $\,6\,$, up to isomorphism, is $\,S_3\,$ ,we get that $\,S_4/Z_4\cong S_3\,$

A coset of $V_4$ is a set $gV_4=\{gv:v\in V_4\}$. So in fact any coset of $V_4$ formed by a $g\in G$ which is not in $V_4$ will contain no elements of $V_4$. (This is easy to prove; try it.)

DonAntonio has shown you a somewhat advanced way to see how $S_3\cong S_4/V_4$, but considering your questions for us, I think that the best way to see it, at this point, would be to write out the actual isomorphism $\mu:S_4/V_4\rightarrow S_3$. Use the cosets they've given you. Ensure you know how coset multiplication works - $(gV_4)(hV_4)=(gh)V_4$. You might have to go back and review some definitions.

Where are the rest of the elements of $S_4/V_4$? Try computing one of them. You'll see that it's equal to one of the cosets already listed in the green section. (Remember that the order of elements in a set doesn't matter, and also that re-cycling the elements in a cycle also doesn't change it: (1,2,4,3) = (2,4,3,1) = (4,3,1,2) = (3,1,2,4)).