Yes. For me, everything is much less confusing if you think of ideals $I$ in terms of the corresponding quotient maps $A \to A/I$. Then the ideals of $f(A)$ are precisely the quotient maps $f(A) \to f(A)/I$, which composed with the map $A \to f(A)$ are precisely those maps $A \to f(A) \to f(A)/I$ which factor through $f(A)$ - in other words, which have kernel containing $\ker(f)$.
Moreover, an ideal $I$ is prime if and only if $A/I$ is an integral domain and maximal if and only if $A/I$ is a field. Hence if $f(A) \to f(A)/I$ has image an integral domain then the composition $A \to f(A) \to f(A)/I$ also has image an integral domain, and the same is true if you replace "integral domain" by "field." Other properties of ideals can also be stated in this way, hence also pull back: for example, an ideal $I$ is radical if and only if $A/I$ has no nilpotents.