Write your function $U(V_1,V_2,W_1,W_2) = V_1 W_2-V_2 W_1$. Then just compute the partial derivatives: $\frac{\partial U(V_1,V_2,W_1,W_2)}{\partial V_1} = W_2$, $\frac{\partial U(V_1,V_2,W_1,W_2)}{\partial V_2} = -W_1$, $\frac{\partial U(V_1,V_2,W_1,W_2)}{\partial W_1} = - V_2$, $\frac{\partial U(V_1,V_2,W_1,W_2)}{\partial W_2} = V_1$. So, the derivative at $(H,K)$ is $\begin{bmatrix} K_2 & -H_2 \\ -K_1 & H_1\end{bmatrix}.$
Note: My answer needs some elaboration.
The derivation is correct, but the expression of the answer needs some elaboration. The derivative is a linear map $\mathbb{R}^2 \times \mathbb{R}^2 \to \mathbb{R}$. To evaluate the derivative in a particular direction ($(H,K)$ in this instance) the above expression must be multiplied componentwise by the corresponding components of the direction.
The derivative of $\det$ evaluated at $(V,W)$ in the direction $(H,K)$ is given by $\frac{\partial U(V_1,V_2,W_1,W_2)}{\partial V_1} H_1+\frac{\partial U(V_1,V_2,W_1,W_2)}{\partial V_2} H_2 +\frac{\partial U(V_1,V_2,W_1,W_2)}{\partial W_1} K_1 +\frac{\partial U(V_1,V_2,W_1,W_2)}{\partial W_2} K_2$, which simplifies to $W_2 H_1-W_1 H_2-V_2 K_1+V_1 K_2 = \det(H,W)+\det(V,K)$.