Suppose $A,B$ are $n\times n$ matrices with coefficients in a field $\mathbb{F}$. Is it always possible to find $n\times n$ matrices $M,N$ with coefficients in the same field such that $$\ker(MA+NB) = \ker(A)\cap \ker(B)?$$ If $\mathbb{F} = \mathbb{R}$ or $\mathbb{C}$ then we can set $M = A^T, N = B^T$ and then the result follows because $$A^TAv + B^TBv = 0 \implies 0 = \langle v, A^TAv + B^TBv\rangle $$ $$= \langle v, A^TAv\rangle + \langle v,B^TBv\rangle = \|Av\|^2 + \|Bv\|^2 $$ which immediately tells us that $Av = Bv = 0$ (where the inner products are the normal dot product and sesquilinear products on $\mathbb{R}^n, \mathbb{C}^n$ depending on if we are working over $\mathbb{R}$ or $\mathbb{C}$).
Notably, this approach catastrophically fails over fields not equal to $\mathbb{R}, \mathbb{C}$ as for example we can have $$A = \begin{pmatrix}1 & 1 \\\ 1 & 1\end{pmatrix} \implies A^TA = 0$$ if you are working over $\mathbb{F}_2$. Is it still possible to find $M,N$ despite the fact that this "obvious guess" fails?
Yes. In principle, since $\ker A\cap\ker B=\ker\pmatrix{A\\ B}$, any matrix $\pmatrix{M&N}\in\mathbb F^{n\times2n}$ that maps the range of $\pmatrix{A\\ B}$ to a subspace of the same dimension will do the job.
E.g., suppose $\pmatrix{A\\ B}$ has rank $r>0$. Let $UV$ be its rank factorisation, where $U\in\mathbb F^{2n\times r}$. Let $T\in\mathbb F^{r\times 2n}$ be a left inverse of $U$. Then $$ \ker\left(\pmatrix{T\\ 0_{(n-r)\times 2n}}UV\right)=\ker(TUV)=\ker(I_rV)=\ker(V)\subseteq\ker(UV)\subseteq\ker(TUV). $$ Therefore all set inclusions above are equalities. Now put $\pmatrix{M&N}=\pmatrix{T\\ 0_{(n-r)\times 2n}}$ and we are done.
(In the example above, since $UV$ is a rank factorisation, the range of $UV$ is the precisely range of $U$. As $T$ is a left inverse of $U$, it maps the range of $U$ to a subspace of the same dimension.)