Consider a differential equation $X' = F(X)$, where $X(t) = (x_1(t),x_2(t))$. This can also be written as
$ \begin{align*} x_1' &= F_1(x_1,x_2), \\ x_2' &= F_2(x_1,x_2), \end{align*} $
where $F = (F_1,F_2)$. Suppose the system has an equilibrium at $X = X_0$. The first step is to compute the linearization about $X_0$, which is given by
$ Y' = D_XF(X_0)\,Y. $
To unravel this notation a bit, suppose $X_0 = (\alpha,\beta)$. Then $D_XF(X_0) = \left( \begin{array}{cc} \frac{\partial F_1}{\partial x_1}(\alpha,\beta) & \frac{\partial F_1}{\partial x_2}(\alpha,\beta) \\ \frac{\partial F_2}{\partial x_1}(\alpha,\beta) & \frac{\partial F_2}{\partial x_2}(\alpha,\beta) \end{array} \right).$ This is just the matrix of the first partial derivatives of the components of $F$ evaluated at the equilibrium point. In your problem, $X_0 = \left(\frac{d}{c},\frac{a}{b}\left(K-\frac{d}{c}\right)\right)$ and $F(f,g) = \left(\begin{array}{c} af(K-f)-bfg \\ cfg-dg \end{array}\right)$ (so that $F_1(f,g) = af(K-f)-bfg$ and $F_2(f,g) = cfg-dg$).
A linear system like this is asymptotically stable at the origin (and hence $X=X_0$ is an asymptotically stable equilibrium of $X' = F(X)$) if both eigenvalues of the matrix $D_XF(X_0)$ have negative real part.
For your problem, it is indeed the case that both eigenvalues have negative real part when $K > d/c$, so the equilibrium in question is indeed asymptotically stable.