Operations on OBDD: negation through Shannon's expansion
Maybe it is better to understand if you make a truth table for both functions. You can see that if $(\neg f|_{x_1=0} \wedge \neg f|_{x_1=1})$ is true then $(\neg x_1 \wedge \neg f|_{x_1=0}) \vee  (x_1 \wedge \neg f|_{x_1=1})$ is also true, because both $f$ terms are true and either $x_1$ is true or $\neg x_1$ is true. And if $(\neg f|_{x_1=0} \wedge \neg f|_{x_1=1})$ is false then you have to do a case study: 


If both $f$ terms are false then is $(\neg x_1 \wedge \neg f|_{x_1=0}) \vee  (x_1 \wedge \neg f|_{x_1=1})$ also false
If exactly one $f$ term (w.l.o.g. $\neg f|_{x_1=0}$) is false then it is equivalent to $(x_1 \wedge \neg f|_{x_1=1})$


Therefore, both formulas $(\neg x_1 \wedge \neg f|_{x_1=0}) \vee  (x_1 \wedge \neg f|_{x_1=1}) \vee (\neg f|_{x_1=0} \wedge \neg f|_{x_1=1})$ and $(\neg x_1 \wedge \neg f|_{x_1=0}) \vee  (x_1 \wedge \neg f|_{x_1=1})$ are equivalent.