Can choice function in SHA algorithm be used in a simplied way?
I'm not familiar with the details of SHA but
if we interpret $0$ as false and $1$ as true, then $a+b$ and $a \oplus b$ only differ when $a=b=1$.
In your specific case $(x \wedge y)$ and $(\neg x \wedge z)$ cannot both be true.
This is easy to see since you would need to simultaneously have $x=1$ and $x=0$.
It follows that  $(x \wedge y) \oplus (\neg x \wedge z)$  and $(x \wedge y) + (\neg x \wedge z)$ are equivalent. The same holds true if $x$, $y$, and $z$ are binary numbers and $\neg$, $\wedge$, and $\oplus$ are interpreted as bitwise operators.
This equality does not hold for the case of $\text{Maj}(x,y,z)$, as you can check by setting $x=y=z=1$.