For given reduction f, can show "if f(x) in 4NAE then x in 3SAT", but not "if x is not in 3SAT then f(x) not in 4NAE"
The error is in this paragraph:


  We know $\varphi \notin \mathsf{3SAT}$ iff any non-contradictory assignment of literals has at least one clause with all false values. However, assuming there is no clause in $$ containing all true literals, then if $()$ adds $=$ to each clause, wouldn't $()$ belong to $\mathsf{4NAE}$?


Here's your argument, broken into steps:


Suppose $\varphi \notin \mathsf{3SAT}$. Thus, every assignment for $\varphi$ has at least one falsified clause.
Suppose there was some assignment in which each clause contains a falsified literal. Then $f(\varphi) \in \mathsf{4NAE}$, by choosing $z=T$.


I fail to see the contradiction here. Indeed, if there is some assignment in which each clause contains a falsified literal, then if you complement it, you get a satisfying assignment, as you show in your post. This means that if (1) holds, then (2) cannot possibly hold.

You argument is basically saying "what if I had an unsatisfiable $\varphi$ together with an assignment in which each clause contains a falsified literal". If such a $\varphi$ existed, then your reduction wouldn't have been sound. Fortunately, no such $\varphi$ exists. It's true that you have to rule out the existence of such $\varphi$, which you do by showing that if (2) holds then (1) doesn't hold.