$\def\rad{\text{rad}}$ One approach is to use your previous problem.
Let $S$ be an arbitrary simple module. We know that
$$\text{Hom}_A(M\oplus N,S)\cong \text{Hom}_A(M,S)\oplus\text{Hom}(N,S)\quad\mathbf{(1)}$$
with the isomorphism sending sending $(f,g)$ in the RHS to $F$ in the left-hand side defined by $F(m,n)=f(m)+g(n)$. Now, from this we see that if $(m,n)\in\rad(M)\oplus\rad(N)$ then $(m,n)\mapsto f(m)+g(n)=0$ for all $(f,g)$ in the RHS of $\mathbf{(1)}$ and thus $F(m,n)=0$ for all $F$ in the LHS of $\mathbf{(1)}$ by the isomorphism.
Conversely, if $(m,n)\in\rad(M\oplus N)$ then for all $F$ in the LHS of $\mathbf{(1)}$ we have that $F(m,n)=0$. But, if $f\in\text{Hom}_A(M,S)$ then $F(m,n):= f(m)$ is an $A$-map $M\oplus N\to S$, and thus $f(m)=F(m,n)=0$. Similarly, for any $g\in\text{Hom}_A(N,S)$ the map $G(m,n):=g(n)$ is an element of the LHS of $\mathbf{(1)}$, and thus $g(n)=G(m,n)=0$.