Assume $A\in\mathbb{R}^{n/2\times n}$ with rank $n/2$ and $B\in\mathbb{R}^{n\times n}$ with rank $n$. Under what conditions can I claim that ${\cal N}(AB)\cap {\cal N}(A)=\{0\}$, where ${\cal N}(\cdot)$ stands for nullspace?
Also, under what conditions can I claim ${\cal N}(AB^m)\cap {\cal N}(A)=\{0\}$ where $m$ is a positive integer number.
Obviously, $B$ cannot be the identity matrix for the first case or an involutory matrix for the latter.
Let $(e_1,\ldots,e_{\tfrac{n}{2}})$ be a basis for $\mathcal{N}(A)$, and extend this to a basis $(e_1,\ldots,e_n)$ for $\Bbb{R}^n$. Then $\mathcal{N}(AB)\cap\mathcal{N}(A)=\{0\}$ if and only if $B$ maps each of $e_1,\ldots,e_{\tfrac{n}{2}}$ outside $\mathcal{N}(A)$. This is the case if and only if the first $\tfrac{n}{2}$ columns of $B$ (with respect to this basis!) are not in $\mathcal{N}(A)$.
An example of a matrix $B$ for which $\mathcal{N}(AB^m)\cap\mathcal{N}(A)=\{0\}$ holds for all positive integers $m$ is $$B=I_n+\begin{pmatrix}0&I_{\tfrac{n}{2}}\\0&0\end{pmatrix}.$$
Also note that, because $B$ is of full rank, we have $\mathcal{N}(AB)=B^{-1}\mathcal{N}(A)$. So you want $$B^{-1}\mathcal{N}(A)\cap\mathcal{N}(A)=\{0\}.$$