The original question was well posed, while the edit made it wrong. Let's clarify things first: the term scale normalized derivative was introduced (to my knowledge) in
Mikolajczyk, K. and Schmid, C. 2001. Indexing based on scale invariant
interest points. In Proceedings of the 8th International
Conference on Computer Vision, Vancouver, Canada, pp. 525–
531.
referring to a derivative of order $m$ of image $I(\mathbf{x})$, obtained by convolving with the corresponding derivative of a Gaussian function, multiplied by the derivation scale at the power of $m$. In formulas this is
$$
D_{i_1 \dots i_m}(\mathbf{x},\sigma_D) = \sigma_D^m L_{i_1 \dots i_m}(\mathbf{x},\sigma_D)
$$
The indexes are generic because of the order is $m$. If you use order $m=1$, then you get just the two first order partial derivatives:
$$
D_x(\mathbf{x},\sigma_D) = \sigma_D L_x(\mathbf{x},\sigma_D)\\
D_y(\mathbf{x},\sigma_D) = \sigma_D L_y(\mathbf{x},\sigma_D)
$$
Why do you multiply by $\sigma_D^m$? Because, in this way, the scale normalized derivatives behave nicely under scaling of the intensity pattern. The derivatives obtained on a scaled image will have the same value of those on another scale but with the right derivation scale, that is a different level on the scale space.
Now, just look at $L_x$, since for the other one is exactly the same. This is defined as:
$$
L_x(\mathbf{x},\sigma_D) = g_x(\sigma_D)*I(\mathbf{x})
$$
In your edited answer you observe that derivation and convolution may be exchanged, and in fact you can write:
$$
g_x(\sigma_D)*I(\mathbf{x}) = g(\sigma_D)*I_x(\mathbf{x})
$$
The problem is that in the Harris technique you introduce a non-linear operation: square and product of the two partial derivatives. So you get:
$$
\left[
\begin{array}{cc}
L_x^2(\mathbf{x},\sigma_D) & L_xL_y(\mathbf{x},\sigma_D)\\
L_xL_y(\mathbf{x},\sigma_D)&L_y^2(\mathbf{x},\sigma_D)
\end{array}
\right]
\neq
g(\sigma_D)*
\left[
\begin{array}{cc}
I_x^2(\mathbf{x},\sigma_D) & I_xI_y(\mathbf{x},\sigma_D)\\
I_xI_y(\mathbf{x},\sigma_D)&I_y^2(\mathbf{x},\sigma_D)
\end{array}
\right]
$$
So, to conclude, the first convolution is used to compute the derivatives (obtaining a representation in the scale space at the derivatin scale), then you need to integrate over a window for the Harris algorithm, with the so called integration scale.
Who introduced the two terms providing a good distinction of the two values? I believe that the first appearance of the terms is in
Mikolajczyk, K., & Schmid, C. (2004). Scale & affine invariant interest point detectors. International Journal of Computer Vision, 60(1), 63-86.
In the paper, they set $\sigma_D = 0.7\sigma_I$. Then the paper begins describing how the Harris-Laplace Detector works, but it's really complex and I still do not understand it very well (10 years after the appearance of the paper...).