Answer
The choice of ${\bf{n}} = {\bf{u}} \times {\bf{v}}$ for the normal vector does not affect the resultant equation.
Work Step by Step
In the derivation of the equation of the tangent plane we chose ${\bf{n}} = {\bf{v}} \times {\bf{u}}$ such that (see page 784)
${\bf{n}} = {\bf{v}} \times {\bf{u}} = \left( {{f_x}\left( {a,b} \right),{f_y}\left( {a,b} \right), - 1} \right)$
Suppose that we choose ${\bf{n}} = {\bf{u}} \times {\bf{v}}$ instead. Then
${\bf{n}} = {\bf{u}} \times {\bf{v}} = \left( { - {f_x}\left( {a,b} \right), - {f_y}\left( {a,b} \right),1} \right)$
With this choice of ${\bf{n}}$ the equation of the tangent plane becomes
$ - {f_x}\left( {a,b} \right)\left( {x - a} \right) - {f_y}\left( {a,b} \right)\left( {y - b} \right) + \left( {z - f\left( {a,b} \right)} \right) = 0$
It follows that
$z = f\left( {a,b} \right) + {f_x}\left( {a,b} \right)\left( {x - a} \right) + {f_y}\left( {a,b} \right)\left( {y - b} \right)$
We conclude that either we choose ${\bf{n}} = {\bf{v}} \times {\bf{u}}$ or ${\bf{n}} = {\bf{u}} \times {\bf{v}}$, we get the same equation of the tangent plane as is given in Theorem 1. Hence, the choice of ${\bf{n}} = {\bf{u}} \times {\bf{v}}$ for the normal vector does not affect the resultant equation.
