MAX-2-XOR-SAT: Why does the special case work?
No, you can't make a substitution like that.  The resulting formula isn't equivalent to the original formula.

Consider

$$(x_1 \oplus x_2) \land (x_1 \oplus \neg x_2).$$

This formula is not satisfiable.  However, if we apply your substitution, we end up with the formula

$$(x_1 \oplus x_2) \land (x_1 \oplus y_2).$$

The resulting formula now has 6 variables and is satisfiable.  For example, you can set $x_1=\text{True}$, $x_2=\text{False}$, $y_2=\text{False}$.

The problem is that your substitution doesn't ensure that $y_i = \neg x_i$, so it doesn't preserve satisfiability.  (If you wanted to preserve satisfiability, you could do that by making your substitution and then adding clauses of the form $(y_i \oplus x_i)$ ... but while that would preserve satisfiability, it wouldn't preserve the maximum number of clauses that can be simultaneously satisfied, so it still wouldn't help for solving MAX-2-XOR-SAT.)