If $A=|t|$, then $A^2 = t^2$; so $\frac{dA^2}{dt} = \frac{d}{dt}t^2 = 2t$ for all $t$.
On the other hand, $\begin{align*} 2A\frac{dA}{dt} &= \left\{\begin{array}{ll} 2|t|&\text{if }t\gt 0;\\ 2|t|(-1)&\text{if }t\lt 0 \end{array}\right.\\ &= 2t,\quad t\neq 0.\end{align*}$ So they are equal where they are both defined, but not equal at $t=0$,as $2A\frac{dA}{dt}$ does not exist there.
For a more radical example, take $A(t) = \left\{\begin{array}{ll}1 &\text{if }t\in\mathbb{Q},\\ -1&\text{if }t\notin\mathbb{Q}. \end{array}\right.$
Then $A(t)$ is not continuous anywhere, so the derivative does not exist anywhere; however, $(A(t))^2 = 1$ for all $t$, so the derivative always exists (and is equal to $0$). Looking at $\frac{d}{dt}A^2 = 2A\frac{dA}{dt},$ the left hand side makes sense, but the right hand side does not (since $\frac{dA}{dt}$ does not exist).
The key point here is that the Product Rule assumes that both factors are differentiable. It is possible for a product to be differentiable and yet for each factor to not be differentiable. In that situation, the Product Rule does not apply.