$\det(A_1\cdot B_1 +A_2\cdot B_2)=0$

If $\det (A_1^2+A_2^2) = 0$, then $x^TA_1^2+x^TA_2^2=0$ for some $x\neq 0$ and so $\|x^TA_1\|^2 + \|x^TA_2\|^2 = 0$.

Since $x^TA_1=x^TA_2 = 0$, we see that $x^T (A_1 B_1+A_2 B_2) =0$ and so $\det (A_1 B_1+A_2 B_2) =0$

Let $\det(A^2_1+A^2_2)=0$ then there exists some vector $x\neq 0$ such that $(A^2_1+A^2_2)x=0$ this implies $$0=\langle (A_1^2+A^2_2)x,x\rangle=\langle A^2_1x,x\rangle+\langle A^2_2x,x\rangle=\langle A_1 x,A^T_1x\rangle+\langle A_2x,A^T_2x\rangle$$ Since $A_1, A_2$ are symmetric then $A_1=A_1^T, A_2=A^T_2$. Therefore $$0=\langle A_1 x,A^T_1x\rangle+\langle A_2x,A^T_2x\rangle=\langle A_1 x,A_1x\rangle+\langle A_2x,A_2x\rangle=||A_1x||^2+||A_2x||^2$$ This holds only if $A_1x=A_2x=0$ in which case we obtain that $\det A_1=\det A_2=0$. Now let $C:=A_1B_1+A_2B_2$ then $C^T=B_1^TA^T_1+B^T_2A^T_2=B_1^TA_1+B^T_2A_2$. This then gives $$C^T x=(B_1^TA_1+B^T_2A_2) x=B_1^TA_1x+B^T_2A_2x=B_1^T0+B^T_20=0$$ Therefore $\det C^T=0$ but $\det C=\det C^T$ for any matrix $C$. The result follows $$\det C=\det (A_1B_1+A_2B_2)=0$$