(1) Given the equation of the parabola $C$ is $x^2=4y$, we can find the equation of the directrix: $y=-1$ and the coordinates of the focus: $F(0,1)$.
(2) Proof: Let the equation of line $AB$ be $y=kx+b$, with $A(x_1,y_1)$, $B(x_2,y_2)$.
By solving the system of equations $\begin{cases} y=kx+b \\ x^2=4y \end{cases}$, we get $x^2-4kx-4b=0$.
Therefore, $\begin{cases} x_1+x_2=4k \\ x_1x_2=-4b \end{cases}$,
$\overrightarrow{OA} \cdot \overrightarrow{OB} = x_1x_2 + y_1y_2 = x_1x_2 + \frac{x_1^2x_2^2}{16} = -4$,
Thus, $x_1x_2=-8$,
Therefore, $-4b=-8$, which gives $b=2$,
The line $y=kx+2$ always passes through the fixed point $(0,2)$.
So, the final answers are:
(1) The equation of the directrix is $\boxed{y=-1}$ and the coordinates of the focus are $\boxed{F(0,1)}$.
(2) The line $AB$ always passes through the fixed point $\boxed{(0,2)}$.