Solution:

We will use the method of finite differences. Define $b_{n} = a_{n + 3} - 3a_{n + 2} + 3a_{n + 1} - a_{n}$. Since $f$ is quadratic, the third finite difference of $f$ is zero. So, $b_{n} = g_{n + 3} - 3g_{n + 2} + 3g_{n + 1} - g_{n}$. Letting the common ratio of the geometric sequence be $r$, we get that $b_{n} = (r^{3} - 3r^{2} + 3r - 1)g_{n}$. So, $b_{n}$ is a constant multiple of $g_{n}$. Thus the ratio $\frac{g_{2}}{g_{1}} = \frac{b_{2}}{b_{1}}$. Computing $b_{1} = 14 - 3\cdot 3 + 3\cdot 2 - 1 = 10$ and $b_{2} = 16 - 3\cdot 14 + 3\cdot 3 - 3\cdot 2 = -19$, we get

$$\frac{g_{2}}{g_{1}} = \frac{b_{2}}{b_{1}} = \boxed{\frac{19}{10}}.$$