For the pedantic's sake, we first have a polynomial reduction $3SAT \leq_p s3SAT$, where the later has strictly 3 terms per clause, not less (as accepted by the former). This is achieved by taking an instance of $3SAT$ and mapping the clauses respectively:
$ (a) \mapsto (\lnot s \lor \lnot s \lor \lnot s) \land (s \lor s \lor a) $
$ (a \lor b) \mapsto (\lnot s \lor \lnot s \lor \lnot s) \land (s \lor b \lor a) $
where $s$ is a dummy variable, and a and b are terms where $\lnot \lnot$ is negated respectively. We concatenate the end result with $\land$. Clearly we construct in polynomial time.
Next we do the polynomial reduction $s3SAT \leq_p NAE-4SAT$, where the second is simply $NAE-3SAT$ but with four terms per clause. For this we take an instance of $s3SAT$ and map each clause as follows:
$ (x \lor y \lor z) \mapsto (x \lor y \lor z \lor s) \land (\lnot x \lor \lnot y \lor \lnot z \lor \lnot s)$
Where $x,y,z$ are terms and $s$ is our dummy variable, as before. Notice the symmetry here, if we find an assignment with $s = true$ we can simply invert the assignment to receive another valid assignment with $s=false$. This is the assignment we want.
As a last step we reduce $NAE-4SAT \leq_p NAE-3SAT$. Take an instance in $NAE-4SAT$ and map the clauses as follows:
$ (a \lor b \lor c \lor d) \mapsto (s \lor a \lor b) \land (\lnot s \lor c \lor d)$
All the same as before, concatenate the result once more. Notice here that if the true and false variable's (one of each must exist) are mapped to the same clause, $s$ can be choose appropriately. If they are mapped to different clauses, $s$ can be choose opposite to the respective variable value in each pair.
To summarize: $3SAT \leq_p s3-SAT \leq_p NAE_4SAT \leq_p NAE-3SAT$.
Answer to extra question: 3SAT only allows $\lor$ in the clauses.