Let's break down the solution into detailed steps:

**Part (1): Finding the Equation of Ellipse $C$**

Given that the right focus is $F(\sqrt{2}, 0)$, we know the semi-focal distance $c = \sqrt{2}$. The eccentricity of the ellipse is given as $e = \frac{\sqrt{6}}{3}$. Since the eccentricity $e = \frac{c}{a}$, we can solve for $a$:

$$e = \frac{c}{a} \Rightarrow \frac{\sqrt{6}}{3} = \frac{\sqrt{2}}{a} \Rightarrow a = \sqrt{3}$$

Next, we find $b^2$ using the relationship $b^2 = a^2 - c^2$:

$$b^2 = (\sqrt{3})^2 - (\sqrt{2})^2 = 3 - 2 = 1$$

Thus, the equation of the ellipse is:

$$\frac{x^2}{3} + y^2 = 1$$

**Part (2): Proving $M$, $N$, and $F$ are Collinear if and Only if $|MN|=\sqrt{3}$**

Proof of Sufficiency:

Assuming the line $MN$ has the equation $y = kx + b$ (where $kb 0$), we have:

$$\frac{|b|}{\sqrt{k^2 + 1}} = 1 \Rightarrow b^2 = k^2 + 1$$

Solving the system of equations $\left\{\begin{array}{l}y = kx + b,\\ \frac{x^2}{3} + y^2 = 1\end{array}\right.$, we get:

$$(1 + 3k^2)x^2 + 6kbx + 3b^2 - 3 = 0$$

For the quadratic equation to have real roots, $\Delta = 36k^2b^2 - 4(1 + 3k^2)(3b^2 - 3) > 0$, which simplifies to $b^2 0$, the sum and product of roots are $x_1 + x_2 = \frac{3\sqrt{2}}{2}$ and $x_1 \cdot x_2 = \frac{3}{4}$, respectively. Thus, $|MN| = \sqrt{3}$, establishing necessity.

Therefore, we conclude that $M$, $N$, and $F$ are collinear if and only if $|MN| = \sqrt{3}$, encapsulated as:

$$\boxed{|MN| = \sqrt{3}}$$