Let's break down the solution step by step, adhering closely to the given solution and formatting the final answers as requested:

### Part (1): Find the equation of the ellipse $C$

Given $C: \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ with eccentricity $e = \frac{\sqrt{3}}{3}$, where $e = \frac{c}{a}$, we can write:
$$e = \frac{c}{a} = \frac{\sqrt{3}}{3}$$
This gives us the relation:
$$a = \sqrt{3}c$$
Given that the minimum distance from a point on the ellipse to the focus is $\sqrt{3} - 1$, we have:
$$a - c = \sqrt{3} - 1$$
Substituting $a = \sqrt{3}c$ into the above equation, we solve for $a$ and $c$:
$$a = \sqrt{3}, \quad c = 1$$
Using $b^{2} = a^{2} - c^{2}$, we find:
$$b^{2} = 3 - 1 = 2$$
Therefore, the equation of the ellipse is:
$$\boxed{\frac{x^{2}}{3} + \frac{y^{2}}{2} = 1}$$

### Part (2): Find the general equation of the line $l$

Given the inclination of the line $l$ is $45^{\circ}$, we can express $l$ as:
$$y = x + m$$
For the intersection points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$, solving the system:
$$\left\{\begin{array}{l} \frac{x^{2}}{3} + \frac{y^{2}}{2} = 1 \\ y = x + m \end{array}\right.$$
leads to the quadratic equation:
$$5x^{2} + 6mx + 3m^{2} - 6 = 0$$
For real solutions, the discriminant $\Delta$ must be positive:
$$\Delta = 36m^{2} - 4 \cdot 5(3m^{2} - 6) = 24(5 - m^{2}) > 0$$
Thus:
$$-\sqrt{5} < m < \sqrt{5}$$
Using Vieta's formulas:
$$x_1 + x_2 = -\frac{6m}{5}, \quad x_1x_2 = \frac{3m^2 - 6}{5}$$
The distance $|AB|$ is given by:
$$|AB| = \sqrt{2} \sqrt{{\left(-\frac{6}{5}m\right)}^2 - 4 \times \frac{3m^{2} - 6}{5}} = \frac{8\sqrt{3}}{5}$$
Solving this equation, we find:
$$m = \pm 1$$
Therefore, the general equations of line $l$ are:
$$\boxed{x - y - 1 = 0} \quad \text{or} \quad \boxed{x - y + 1 = 0}$$