Suppose I have to integrate $dy \wedge dz$ over the unit hemisphere centered at the origin. I have:
$$ \int_S dy \wedge dz$$
The equation of hemisphere is given as $z=\sqrt{1-x^2-y^2}$, I have $dz = \frac{\partial z}{\partial x} dx + \frac{\partial z}{\partial y } dy$. I can substitute this expression of $dz$ in the above to get:
$$ \int_S \frac{\partial z}{\partial x}dy \wedge dx $$
Conceptually, what exactly happened in the above substitution? I am a bit confused because $dy \wedge dz$ is an area element in the $y-z$ plane and $dy \wedge dx$ is an area element in the $x-y$ plane. Note that I'm using the interpretation of two form as something you project the area spanned by two vectors into.
The chosen coordinate chart for the upper hemisphere $S_+$ of the unit sphere $S$ is $$\tag{1} \{x^2+y^2\le 1\}\ni (x,y)\mapsto \begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}x\\y\\\sqrt{1-x^2-y^2}\end{pmatrix}\in S_+. $$ Intuitively and geometrically the $2$-form $dx\wedge dy$ is an oriented surface element that is spanned by the tangent vectors on $S$ in the $x$- and $y$-direction. In $\mathbb R^3$ we know that this can be expressed by a cross product $$ \begin{pmatrix}\frac{\partial x}{\partial x}\\ \frac{\partial y}{\partial x}\\\frac{\partial z}{\partial x} \end{pmatrix}\times\begin{pmatrix}\frac{\partial x}{\partial y}\\ \frac{\partial y}{\partial y}\\\frac{\partial z}{\partial y} \end{pmatrix} =\begin{pmatrix}1\\0\\\frac{-x}{z}\end{pmatrix}\times \begin{pmatrix}0\\1\\\frac{-y}{z}\end{pmatrix}= \frac{1}{z}\begin{pmatrix}x\\y\\z\end{pmatrix}\,. $$ Then we have a vector valued surface integral $$\tag{2} \int_{S_+}dx\wedge dy=\int_{x^2+y^2\le 1}\frac{1}{z}\begin{pmatrix}x\\y\\z\end{pmatrix}\,dx\,dy. $$ The integration variables $x,y$ on the RHS of this formula are ordinary numbers.
When you want to consider $$ \int_{S_+}dy\wedge dz $$ I think you should use the chart $$\tag{2} \{y^2+z^2\le 1\}\ni (y,z)\mapsto \begin{pmatrix}x\\y\\z \end{pmatrix}=\begin{pmatrix}\sqrt{1-y^2-z^2}\\y\\z\end{pmatrix}\in S_+. $$ and repeat the whole exercise. But note that the hemisphere $S_+$ is now a different one.