(1) The equation of circle $C$ can be rewritten as $(x^2)+((y-4)^2) = 16$. Therefore, the center of the circle is $C(0,4)$ and its radius is $4$.

The chord $AB$ is shortest when it is perpendicular to $MC$. At that time, the length of chord $AB$ is $2\sqrt{R^2 - CP^2} = 4\sqrt{2}$, and the equation of line $l$ satisfying this condition is $x-2y+2=0$.

To elaborate, let us consider that the shortest chord through point $P$ will be the chord that is perpendicular to the radius of the circle through point $P$. In this case, line $l$ will pass through both $P(2,2)$ and the closest point to $P$ on the circle along the direction of the radius. That point can be found from the circle's equation to be $C(0,4)$. Since the circle has a radius of 4, and using the fact that $P$ lies inside the circle, we can use the Pythagorean theorem to find the shortest distance from the center $C$ to $P$, which will be $CP=\sqrt{(2-0)^2+(2-4)^2} = 2\sqrt{2}$. Hence, the shortest length of chord AB is $2\sqrt{R^2 - CP^2} = 2\sqrt{4^2 - (2\sqrt{2})^2} = 4\sqrt{2}$.

(2) Let $M(x,y)$ be the midpoint of chord $AB$. Thus, $\vec{CM} = (x, y-4)$ and $\vec{MP} = (2-x, 2-y)$.

According to the problem, $\vec{CM} \cdot \vec{MP} = 0$ since they are perpendicular when $AB$ is shortest, which gives us the equation
$$x(2-x) + (y-4)(2-y) = 0$$
Expanding and simplifying, we get
$$(x-1)^2 + (y-3)^2 = 2$$

Since point $P$ lies inside the circle $C$, the trajectory of $M$ will be an ellipse with the equation
$$\boxed{(x-1)^2 + (y-3)^2 = 2}$$