How to change basis for normal vector fields on regular surfaces? I have two normal vector fields over a surface $S$, to two parameterizations $\psi,\varphi$:

$$X_{\psi}:p\mapsto X_{\psi}(p)=\dfrac{\partial \psi}{\partial u}(\psi^{-1}(p))\wedge \dfrac{\partial \psi}{\partial v}(\psi^{-1}(p))$$
$$X_{\varphi}:p\mapsto X_{\varphi}(p)=\dfrac{\partial \varphi}{\partial u}(\varphi^{-1}(p))\wedge \dfrac{\partial \varphi}{\partial v}(\varphi^{-1}(p))$$

At the intersection of the neighborhoods, $X_\psi,X_\varphi$ are related by matrix of coordinate changing, $g=\varphi^{-1}\circ\psi$. A professor wrote:

$$X_{\psi}(q)=\det(Jg(\psi^{-1}(q)))X_{\varphi}(q).$$

I am confused with the transformations: I thought $g$ leads $Dom(\psi)$ to $Dom(\varphi)$, so why is not

$$X_{\fbox{$\varphi$}}(q)=\det(Jg(\psi^{-1}(q)))X_{\fbox{$\psi$}}(q)?$$

Many thanks in advance for a light. If possible, I'd like not a yes/no answer, but some light to deal with this kind of transformation. Please take some look at the comments bellow. Thank you so much!
 A: Consider a function $h(x) = f(u)= f(g(x))$ where $u = g(x)$. Now, it is straightforward to see that $$\frac{\partial h}{\partial x_i} = \frac{\partial h}{\partial u_j} \frac{\partial u_j}{\partial x_j} $$
 Notice that, while changing the variables from $x$ to $u$, the variables appear in the denominator i.e. $\frac{\partial h}{\partial x_i}$ and $\frac{\partial h}{\partial u_j}$.
Now coming to your question, $g=\varphi^{-1}\circ\psi$ and we are rather looking for how $\frac{\partial \varphi}{\partial u}$ and  $\frac{\partial \psi}{\partial u}$ are related. Notice that, unlike the previous case, the 'change of co-ordinates' happen in the numerator. Hence, it is the reverse of what you expected. 
Update:
\begin{align}
\frac{\partial \psi}{\partial  u} (\psi^{-1}(q)) & = \psi^{\prime}(\psi^{-1}(q))\times \frac{\partial \psi^{-1}(q)}{\partial  u} \\
 & = \psi^{\prime}(\psi^{-1}(q))\times \frac{\partial \psi^{-1}(\varphi (\varphi^{-1} (q)))}{\partial  u} \\
& = \psi^{\prime}(\psi^{-1}(q))\times {(\psi ^{-1})}^{\prime}(q) \frac{\partial (\varphi (\varphi^{-1} (q)))}{\partial  u} \\
& = \psi^{\prime}(\psi^{-1}(q)) \times \frac{1}{\psi^{\prime}(\psi^{-1}(q))} \frac{\partial (\varphi (\varphi^{-1} (q)))}{\partial  u} \qquad \text{(from the derivative of inverse function formula)} \\
& = \frac{\partial (\varphi (\varphi^{-1} (q)))}{\partial  u}
\end{align}
I am unable to get the multiplicative Jacobian factor in the final answer. May be someone can help.  
