When does the magnitude of the gradient equal the surface area of the $dxdy$ patch?

Given a surface $$S$$ in $$\mathbb R^3$$, what is the relationship between the gradient (when $$S$$ is defined as a level curve of function $$F: \mathbb R^3 \to \mathbb R$$) and surface area?

I noticed such a relationship when examining the surface area of the saddle surface $$z = x^2 - y^2$$. The surface area has integrand $$\sqrt{1 + 4x^2 + 4y^2}$$. And, when defined as a level curve $$F(x,y,z) = x^2 - y^2 - z$$, we get $$\|\nabla F\| = \sqrt{1 + 4x^2 + 4y^2}$$.

What is the relationship between surface area and the gradient? Does it hold for $$\mathbb R^n$$ or only $$\mathbb R^3$$?

My (incomplete) attempt is below.

Partial Solution

If $$S$$ is a surface defined as the level curve of $$F(x,y,z)$$, and $$\frac {\partial F}{\partial z} = \pm 1$$, then the magnitude of $$\nabla F$$ equals the ratio of surface area of a portion of $$S$$ to its projection on the $$xy$$ plane, and $$\text{Surface Area } = \iint \|\nabla F\| dx dy.$$

Given a surface $$S$$, point $$v \in S$$, and plane $$P$$, define surface amplification of $$S$$ over $$P$$ as $$\lim_{\delta \to 0} \frac{\text{surface area of }R}{\text{area of }R \text{ projected onto } P}$$ where $$R$$ is a region of $$S$$ including point $$v$$ with area $$\delta$$. (Formalizing this limit is hard.)

If $$S$$ is parameterized by $$(x, y, Z(x,y))$$, then surface amplification over the $$xy$$ plane equals $$\sqrt{1 + \frac{\partial Z}{\partial x}^2 + \frac{\partial Z}{\partial y}^2}$$ as can be easily seen from the standard surface area integral.

If $$S$$ is defined as the level surface of $$F(x,y,z)$$, and $$\frac {\partial F}{\partial z} = \pm 1$$, then surface amplification over the $$xy$$ plane equals $$\| \nabla F \|$$, which follows from the definition of gradient. (How do I show this clearly?)

I conjecture this applies for any $$\mathbb R^n$$. This post may be related.

Is this correct? How do we complete the proof? What is the geometric intuition behind this?

Quoting from https://math.stackexchange.com/a/4651813/73934, the volume form of the ambient space $$dx\wedge dy\wedge dz$$, given a normal vector field to a surface, induces the following surface area form:
$$ds=n_{x}\ dy\wedge dz+n_{y}\ dz\wedge dx+n_{z}\ dx\wedge dy=\dfrac{dy\wedge dz}{n_{x}}=\dfrac{dz\wedge dx}{n_{y}}=\dfrac{dx\wedge dy}{n_{z}}$$
The last expression is particularly useful when $$F(x,y,z) = z-f(x,y)$$ because then $$n_z\equiv\frac{1}{\|\nabla F\|}$$ and $$ds=\|\nabla F\|dx\wedge dy$$
The geometric intuition may be that the gradient shows by how much a surface patch on a $$F=const$$ has been stretched compared to its $$xy$$ projection. Looking for intuition it helps to reduce the dimensionality. Let $$F(x,y)=y-f(x)$$. An infinitesimal slope length over $$dx$$ is $$\sqrt{ dx^2 + (f_xdx)^2}=\|\nabla F\|dx.$$