A curious determinantal identity

Once you distill away all the multilinear algebra and restrict to linear transformations $\mathbb{R}^2 \to \mathbb{R}^2$, one can define the adjugate or classical adjoint of a matrix $A \in \mathbb{R}^{2 \times 2}$ to be the unique matrix $\operatorname{adj}(A) \in \mathbb{R}^{2 \times 2}$ such that $$ \forall v_1,v_2 \in \mathbb{R}^2, \quad \det(A v_1 \vert v_2) = \det(v_1 \vert \operatorname{adj}(A)v_2), $$ where for $v_1,v_2 \in \mathbb{R}^2$, $(v_1 \vert v_2)$ denotes the $2 \times 2$ matrix whose first column is $v_1$ and whose second column is $v_2$; a direct computation then shows that $$ \operatorname{adj}\begin{pmatrix}a&b\\c&d\end{pmatrix} = \begin{pmatrix}d&-b\\-c&a\end{pmatrix}. $$ The defining property of $\operatorname{adj}(A)$ immediately implies the following facts:

1. for every $A \in \mathbb{R}^{2 \times 2}$, $\operatorname{adj}(A)A = \det(A)I_2$ (the well-known general property of adjugates);
2. the assignment $A \mapsto \operatorname{adj}(A)$ defines a linear transformation $\operatorname{adj} : \mathbb{R}^{2 \times 2} \to \mathbb{R}^{2 \times 2}$ (a property specific to the $2 \times 2$ case);

which allow us to directly compute $$ \begin{align} &\left(\det\left(\tfrac{1}{2}(A+A^t)\right)+\det\left(\tfrac{1}{2}(A-A^t)\right)\right)I_2\\ =\, &\operatorname{adj}\left(\tfrac{1}{2}(A+A^t)\right)\left(\tfrac{1}{2}(A+A^t)\right) + \operatorname{adj}\left(\tfrac{1}{2}(A-A^t)\right)\left(\tfrac{1}{2}(A-A^t)\right)\\ =\, &\frac{1}{4}\left(\operatorname{adj}(A)A+\operatorname{adj}(A)A^t+\operatorname{adj}(A^t)A + \operatorname{adj}(A^t)A^t+\operatorname{adj}(A)A-\operatorname{adj}(A)A^t-\operatorname{adj}(A^t)A + \operatorname{adj}(A^t)A^t\right)\\ =\, &\frac{1}{2}\left(\operatorname{adj}(A)A+ \operatorname{adj}(A^t)A^t\right)\\ =\, &\frac{1}{2}\left(\det(A)+\det(A^t)\right)I_2\\ =\, &\det(A)I_2. \end{align} $$ So, at the end of the day, as loup blanc's answer points out, what you're really seeing here is the special fact that the operator of taking the adjugate of a square matrix is actually linear in the case of $2 \times 2$ matrices; if you're comfortable with multilinear algebra, then this really boils down to the numerological coincidence that $\wedge^{n-1}\mathbb{R}^n = \mathbb{R}^n$ when $n=2$ (and not just $\wedge^{n-1}\mathbb{R}^n \cong \mathbb{R}^n$, which holds in general).

The relation $\det(A)=\det(B)+\det(C)$ is true when $n=2$ because the application $A\in M_2\rightarrow adj(A)\in M_2$ is linear ($adj(A)$ is the classical adjoint of $A$).

Proof. Let $f:A\in M_2\rightarrow \det(A)-\det(B)-\det(C)$; note that $f=0$ iff, for every $A,H\in M_2$, $Df_A(H)=0$. One has

$Df_A(H)=tr(Hadj(A)-1/4Hadj(A+A^T)-1/4H^Tadj(A+A^T)-1/4Hadj(A-A^T)$

$+1/4H^Tadj(A-A^T))=tr(H(adj(A)-1/2adj(A+A^T)-1/2adj(A-A^T)))=0$.

That is equivalent to $2adj(A)-adj(A+A^T)-adj(A-A^T)=0$. We conclude thanks to the linearity of $adj(.)$.