Why $z = \{ Potato, \{ Potato \} \}$ is not a transitive set ? We need to check this only:
If $y \in z$ and if $x \in y$, then $x \in z$.
$y$ can be either $Potato$ (in this case $x$ is a urelement, so $S = \{ x | x \in y \}$ is empty so the statement if $x \in y$, then $x \in z$ is vacaously true). In the other case $y = \{ Potato \}$, and in this case the only possible contender for $x$ is $Potato$, so $x \in z$.
If this is a transitive set, then how the ordinals is well ordered or even ordered under the order $\in$ ? Because since $z$ is transitive, it lies in the class of all ordinals, as does $\{ \{ \}, \{ \{ \} \} \}$. Then how you can compare $\{ Potato, \{ Potato \} \}$ with $\{ \{ \}, \{ \{ \} \} \}$ ?
In the context of ZF(C), everything is a set. That is, you only can have the set $z=\{\text{Potato},\{\text{Potato}\}\}$ if $\text{Potato}$ itself is a set. But then for that set to be transitive, every element of $\text{Potato}$ has to be an element of $z$. But $z$ has only two elements, namely $\text{Potato}$ and $\{\text{Potato}\}$, which both cannot be elements of $\text{Potato}$ due to the axiom of foundation.
Thus in the context of ZF(C), $z$ is a transitive set if and only if $\text{Potato}=\emptyset$.
If $\text{Potato}$ is meant to be an urelement (that is, an element that is not a set), then you are outside of ZF(C), and need to use a definition of transitive sets that is valid in set theories with urelements.