Note that if $f \in L^q(A)$ and $g \in L^q(B)$, $f(x) g(y) \in L^q(A \times B)$. Then if $k \in L^p(A \times A)$ with $1/p + 1/q = 1$, $\left| \int_{A \times B} k(x,y) f(x) g(y)\ dx \ dy \right| \le \|k\|_p \|f\|_q \|g\|_q$ Since this is true for every $g \in L^q(B)$, $\int k(x,y) f(x)\ dx \in L^p(B)$ with $\|\int k(x,y) f(x)\ dx \|_p \le \|k\|_p \|f\|_q$.
However, it won't work if $r \ne p$. In order to have $\int_A k(x,y) f(x)\ dx \in L^r(B)$, what you'd want is $k(\cdot,y) \in L^p$ for almost all $y \in B$ with $y \to \|k(\cdot,y)\|_p \in L^r(B)$. If $L^p(B)$ is not a subset of $L^r(B)$, this won't be true. In fact, given $h \in L^p(B) \backslash L^r(B)$ and $f\ne 0 \in L^q(A)$, take $g \in L^p(A)$ with $\|g\|_p = 1$ and $\int_A f g \ dx = \|f\|_q$, and define $k(x,y) = g(x) h(y)$. Then $k \in L^p$ with $\|k\|_p = \|g\|_p \|h\|_p$, and $\int_A k(x,y) f(x)\ dx = \|f\|_q h(y) \notin L^r(B)$.