Preserving a propositional formula
In your example the assignment $f_1\wedge f_2$ is $(f_1\wedge f_2)(A)=f_1(A)\wedge f_2(A)=\text{True} \land \text{False} = \text{False}$.
similarly: $(f_1\wedge f_2)(B)=(f_1\wedge f_2)(C)=\text{False}$ so actually $f_1\wedge f_2 \equiv f_2$.

As they define, AND is preserving $X$ if for every two assignments $f_1, f_2$ that satisfy $X$, the assignment $f_1 \wedge f_2$ (defined by $(f_1\wedge f_2)(v)=f_1(v)\wedge f_2(v)$) also satisfies $X$.