I have always been confused about the antisymmetry of the wave function for fermions. Suppose we have a wave function such $\psi$ such that $$\psi(x_1,x_2)=-\psi(x_2,x_1)\tag{1}$$
Now suppose we have coordinate transformation $f_1,f_2$ such that we have
$$\psi(x_1,x_2)= \psi(f_1(y_1,y_1),f_2(y_1,y_2))=\phi(y_1,y_2)\tag{2}$$
How can we prove that
$$\phi(y_1,y_2)=-\phi(y_2,y_1)~?\tag{3}$$
How can we prove that $\phi(y_1,y_2) = -\phi(y_2,y_1)$?
You cannot. Consider $$f_1(y_1,y_2) = \frac{1}{2}(y_1 + y_2),\qquad f_2(y_1,y_2) = \frac{1}{2}(y_1 - y_2).$$ Then, let us take $x_2=0$. In general, $\psi(x_1,0)\neq 0$. We have $$\phi(x_1,x_1) = \psi(x_1,0) \neq 0.$$
The crucial point is that, in the definition of $\psi(x_1,x_2)$, $x_1$ and $x_2$ denote positions of two identical fermions, and this is the reason for $\psi$ to be antisymmetric under the $x_1\leftrightarrow x_2$ transformation.
After the coordinate transformation, $y_1$ and $y_2$ do not necessarily correspond to positions of two fermions. In particular, interchanging the positions of the two fermions doesn't mean that $y_1$ becomes $y_2$ and vice versa. Hence, there is no reason for $\phi$ to be antisymmetric under $y_1\leftrightarrow y_2$.
By your logic you should get $$ f(y,z) = -f(z,y) \tag{1} $$ In any case, the wavefunction under particle interchange is antisymmetric, the individual electrons are identical. Their properties remain the same. We've only interchanged them, not altered them.
