### Solution:

#### Part 1: Finding the Range of $f(x)$

Given the function $f(x) = 2\sqrt{3}\sin x \cos x - 2\cos^{2}x$, we can simplify it using trigonometric identities:

1. Use the double angle formula for sine:
$2\sin x \cos x = \sin 2x$, so $2\sqrt{3}\sin x \cos x = \sqrt{3}\sin 2x$.

2. **Express $\cos^2 x$ using the cosine double angle formula:**
$\cos^2 x = \frac{1 + \cos 2x}{2}$, so $-2\cos^2 x = -1 - \cos 2x$.

3. Combine the results:
$f(x) = \sqrt{3}\sin 2x - 1 - \cos 2x$.

4. Use the sine addition formula:
$\sin(A + B) = \sin A \cos B + \cos A \sin B$, with $A = 2x$ and $B = -\frac{\pi}{6}$, we get $f(x) = 2\sin\left(2x-\frac{\pi}{6}\right) - 1$.

Since $\sin \left(2x-\frac{\pi}{6}\right) \in \left[-1,1\right]$, the range of $f(x)$ is obtained by substituting the maximum and minimum values of the sine function into $f(x)$:

- Maximum: $2(1) - 1 = 1$
- Minimum: $2(-1) - 1 = -3$

Therefore, the range of $f(x)$ is $\left[-3,1\right]$.
$\boxed{\text{The range of } f(x) \text{ is } [-3,1]}$.

#### Part 2: Maximum Area of Triangle $ABC$

Given $f(A) = -2$ and $a = \sqrt{3}$:

1. **Substitute $f(A)$ into the simplified form:**
$2\sin\left(2A-\frac{\pi}{6}\right) - 1 = -2$ implies $\sin\left(2A-\frac{\pi}{6}\right) = -\frac{1}{2}$.

2. **Solve for $A$:**
Since $A \in (0, \pi)$, and considering the range of $2A-\frac{\pi}{6}$, we find $2A-\frac{\pi}{6} = \frac{7\pi}{6}$, which gives $A = \frac{2\pi}{3}$.

3. Use the Law of Cosines:
$a^2 = b^2 + c^2 - 2bc\cos A = b^2 + c^2 + bc$ (since $\cos\frac{2\pi}{3} = -\frac{1}{2}$), and $a^2 = 3$.

4. Find the condition for maximum area:
$3 = b^2 + c^2 + bc \geq 3bc$ implies $bc \leq 1$. The equality holds when $b = c = 1$.

5. Calculate the maximum area:
$S_{\triangle ABC} = \frac{1}{2}bc\sin A \leq \frac{1}{2} \times 1 \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4}$.

Therefore, the maximum value of the area $S$ of triangle $ABC$ is $\boxed{\frac{\sqrt{3}}{4}}$, which occurs when $b = c = 1$.