Maybe a little sketch maybe helpful here.
Consider a one dimensional case first
$$
\ln y = \alpha_0 + \alpha_1 \ln x + \frac{1}{2}\beta_{11}\ln^2x \tag{1}
$$
At first order (that is, ignore $\ln^2 x$), note that
$$
\frac{{\rm d}\ln y}{{\rm d}\ln x} = \alpha_1 \tag{2}
$$
i.e. $\alpha_1$ is just the logarithmic slope of the output function (a.k.a elasticity), so it basically can help you identify the difference between the two plots below

Now the second term, for that one note that
$$
\frac{{\rm d}^2\ln y}{{\rm d}\ln^2 x} = \beta_{11} \tag{3}
$$
there's no mystery involved in the factor $2$ in Eq. (1), just there to make things simpler. Again, this is just the second derivative of the output function, so it can help you tell the difference between the two cases below

Now comes the trick,
How does this extend to higher dimensions?
Well Eq.(1) still holds
$$
\alpha_k = \frac{\partial \ln y}{\partial \ln x_k}
$$
so $\alpha_k$ is the elasticity keeping all other factors constant. Or in geometric terms $\boldsymbol{\alpha} = \nabla_{\ln {\bf x}} \ln y$. An the second derivatives (Hessian) are simply $\beta_{jk}$
$$
\frac{\partial^2 \ln y}{\partial \ln x_k \partial \ln x_l} = \frac{1}{2}(\beta_{kl} + \beta_{lk})
$$
If the matrix with entries $\beta_{kl}$ is diagonal, then the interpretation above is the same, it is just expressing the convexity of the production. If it is not, then you need to look for the eigenvalues of the matrix
$$
\boldsymbol{H} = \frac{1}{2}(\boldsymbol{\beta} + \boldsymbol{\beta}^T)
$$
which always exist and are real, since $\boldsymbol{H} = \boldsymbol{H}^T$