Meaning of the identity $\det(A+B)+\text{tr}(AB) = \det(A)+\det(B) + \text{tr}(A)\text{tr}(B)$ (in dimension $2$)

In a quite pedestrian way, this is just saying that $\det$ is a quadratic form with the trace as its polar form. Namely, for any matrix $A$, let $A^\star = \mathrm{Tr} A - A$ be its conjugate (as in “the conjugate root of the characteristic polynomial”). Then your formula is equivalent to $$ \det (A+B) - \det(A) - \det(B) = \mathrm{Tr}(A^\star \cdot B).$$ The left-hand side is the polar form associated to the determinant form.

In a slightly more abstract way, for any field $k$, the matrix algebra $k^{2\times 2}$ is a (split) quaternion algebra. The formula above is just the polar relation for the norm form in this algebra (which is the determinant of a matrix).

This is the case $n=2$ of a 1980 theorem of Amitsur. It is described (in more abstract notation) here and this question gives the reference to Amitsur's paper.