Let $A$ and $B$ be $n×n$ matrices. If we expand the determinant of $A+B$ as a sum over all permutations of $[\![1,n]\!]$ and all choices of whether the coefficient comes from $A$ or from $B$, this gives a sum with $2^n n!$ terms. We can rearrange it to give a sum over all minors of $A$ of this minor, times the "complementary" minor in $B$, times a sign.
It is a bit hard to explain so I made a program (in sage): https://pastebin.com/qxVRbQMG.
But this formula can be expressed more abstractly. We remark that the $k×k$ minors of $A$ are the coefficients of $\Lambda^k A$ (the exterior power) in the canonical basis. Using the perfect pairing $\Lambda^k E ⊗ \Lambda^{n-k} E → \Lambda^n E$, given an endomorphism $f$ of $\Lambda^k E$, we can define its adjoint $f^* : \Lambda^{n-k} E → \Lambda^{n-k} E$ relatively to this pairing. Given an endomorphism $f$ of $\Lambda^k E$ and an endomorphism $g$ of $\Lambda^{n-k} E$, we define $\langle f, g \rangle$ as $\operatorname{tr}(f^* g)$ (this is again a perfect pairing). Then, the formula translates as $$\det(A+B) = \sum_{k=0}^n \left\langle \Lambda^k A, \Lambda^{n-k} B \right\rangle\text{.}$$
Question: Is there an abstract proof (coordinate-free) of this identity? An intuition behind it? Can we generalize it if it helps?