The two distinct non-zero common roots of $f$ and $g$ are also roots of $g - f$. These are the two roots of the quadratic
$$g(x) - f(x) = 3 \left( x^2 - x - \frac{\beta}{3} - 2 \right),$$
hence we must have
$$\begin{aligned} g(x) &= (x - \lambda) \left( x^2 - x - \frac{\beta}{3} - 2 \right) \\ &= x^3 - (1 + \lambda)x^2 + \left( \lambda - 2 - \frac{\beta}{3} \right)x + \lambda \left( 2 + \frac{\beta}{3} \right) \end{aligned}$$
for some real number $\lambda$. Using that $g(x) = x^3 + (\alpha - 3)x - 6$ and comparing coefficients gives $\lambda = -1$, $\alpha = -\frac{\beta}{3}$ and $\beta = 12$, and hence $\alpha = -4$.