Given two $2 \times 2$ matrices $A$ and $B$,
$$\det(A + B) = 2(\det(A) + \det(B)) - \det(A - B)$$
This can be proven directly via the formula for the determinant of a $2 \times 2$ matrix, and therefore applies whether the matrices are invertible or not.
If $A = \begin{bmatrix}a&c\\b&d\end{bmatrix}$ and
$B = \begin{bmatrix}e&g\\f&h\end{bmatrix}$,
$$\text{tr}(\text{adj}(A)B) = \text{tr}\left(\begin{bmatrix}d&-c\\-b&a\end{bmatrix} \begin{bmatrix}e&g\\f&h\end{bmatrix}\right) = \text{tr}\left(\begin{bmatrix}de-cf&dg-ch\\af-be&ah-bg\end{bmatrix}\right) = de-cf+ah-bg$$
$$\det(A) + \det(B) - \det(A-B) = ad - bc + eh - fg - (a-e)(d-h) + (b-f)(c-g)\\$$
Combining like terms shows that $$\text{tr}(\text{adj}(A)B) = \det(A) + \det(B) - \det(A - B)$$
and therefore that $$\det(A) + \det(B) + \text{tr}(\text{adj}(A)B) = 2(\det(A) + \det(B)) - \det(A - B) = \det(A + B)$$
To answer your other questions, I don't know the name of either formula (I discovered the one I used independently, though I doubt I'm the first). And like I said, my formula applies to all $2\times2$ matrices, invertible or not, and your formula is equivalent to mine, so the matrices need not be of a special form.