We know that (see for example "Vector and Geometric Calculus" by Alan Macdonald, pp. 74, problem 5.4.2) given the covariant basis vectors $\mathbf{b}_1$ and $\mathbf{b}_2$ of the tangential plane $T_P$ of a surface $S$ (defined in $\mathbb{R}^3$) the contravariant basis vectors $\mathbf{b}^1$, $\mathbf{b}^2$ are given by, \begin{equation} \label{one} \mathbf{b}^1 = \frac{\mathbf{b}_2\times(\mathbf{b}_1 \times \mathbf{b}_2)}{|\mathbf{b}_1 \times \mathbf{b}_2|^2} \tag{1.1} \end{equation} and, \begin{equation} \label{two} \mathbf{b}^2 = -\frac{\mathbf{b}_1\times(\mathbf{b}_1 \times \mathbf{b}_2)}{|\mathbf{b}_1 \times \mathbf{b}_2|^2} \tag{1.2} \end{equation} where I have replaced the geometric and outer products with cross products. Given a parametrization, \begin{equation} S \; : \; x^i=x^i(u^1,u^2) \quad , \quad \text{for $i=1,2,3$} \end{equation} then $\mathbf{b}_a=\frac{\partial x^i}{\partial u^a}\rvert_P$ where $a=1,2$.
We also know that the contravariant basis vectors are given by, \begin{equation} \label{three} \mathbf{b}^a = \mathbf{\nabla} u^a \tag{2} \end{equation} (see "Introduction to Differential Geometry of Space Curves and Surfaces", Taha Sochi, p.96, (181)) where $\mathbf{\nabla}=\partial/\partial x^i$.
Is there a way to derive (\ref{one}), (\ref{two}) from (\ref{three})? Given a specific parametrization of the surface (see below) how can we derive the contravariant basis vectors using (\ref{three}) instead of (\ref{one}),(\ref{two})? \begin{align} x^1 & = \cos\theta u^1 - \sin\theta u^2 \\ x^2 & = \sin\theta u^1 + \cos\theta u^2 \\ x^3 & = U + A \exp\left(-\frac{(u^1)^2+(u^2)^2}{2\sigma^2}\right) \end{align}