Since angle $\alpha$ is in the third quadrant, both sine and cosine are negative. Given $\sin \alpha = -\frac{12}{13}$, we can find $\cos \alpha$ using the Pythagorean identity:

$$\cos^2 \alpha + \sin^2 \alpha = 1$$
$$\cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \left(-\frac{12}{13}\right)^2 = 1 - \frac{144}{169} = \frac{25}{169}$$
$$\cos \alpha = -\sqrt{\frac{25}{169}} = -\frac{5}{13}$$

Now we can find $\tan \alpha$ using the definition $\tan \alpha = \frac{\sin \alpha}{\cos \alpha}$:

$$\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{-\frac{12}{13}}{-\frac{5}{13}} = \boxed{\frac{12}{5}}$$

Hence, the answer is B: $\frac{12}{5}$.

This problem primarily tests the understanding of basic trigonometric identities and their applications to simplify trigonometric expressions.