Moment of inertia: why $\mathrm{dI}=r^2\mathrm{dm}$ instead of $\mathrm{dI}=m\mathrm{dr^2}$?

If the body were a discrete set of point mass objects rotating around an axis, you would write $I=\sum_j m_j r_j^2$

For a continuous body the sum goes over into an integral. You consider it as smaller and smaller mass objects. You think of, say, a brick as being made up of grains of rock, then atoms... The objects get smaller $m \to \delta m$ but their radii do not. If the brick has $r=1 m$ then the grains that comprise the brick have $r$ in the range $0.9 - 1.1 m$. So you consider lots of small masses at non-small radii, not the other way round.

Conceptually, each little piece of an object contributes to the total moment of inertia. Each little piece of an object consists of a volume of matter so that each piece has some small mass, $$\delta m = \rho\left(\vec{r}\right) \delta V.$$ $\rho\left(\vec{r}\right)$ is a density function whose value depends on the position of each specific $\delta m$.

Without getting into tensor calculations, i.e., keeping things simplistic, each $\delta m$ has a contribution to the moment of inertia about some axis, $$\delta I = \delta m R_{\perp}^2$$ where $R_{\perp}$ is the perpendicular distance from the axis of interest. (It's actually a little more complicated than that, but the difference is not important to this question.) $R_{\perp}$ is a certain value depending on each $\delta m$. The sum you want to do considers each $\delta m$, not each $R_{\perp}$. (In certain specific symmetries, the sum can be simplified, but you asked about the conceptual reason for summing over the masses.)

As you shrink the volume $\delta V$ to an infinitesimal d$V$, the mass $\delta m$ becomes d$m$, but the specific distance from the axis $R_{\perp}$ doesn't change at all! And so, $$\delta I \to \text{d}I = \text{d}m R_{\perp}^2 = \text{d}V \rho\left(\vec{r}\right) R_{\perp}^2.$$

Integrating this over the complete volume adds the individual contributions of each little volume of matter.

Answer: For a system of continuously distributed total mass $m$, $r$ is constant for a differential (point) mass $dm$ so we use $dI=r^2dm$ but mass $m$ is not constant over a differential distance $dr$ so we don't use $dI=md(r^2)$

Explanation:

When a system of continuously distributed (total) mass $m$ rotates about a point then each point of the system rotates at a certain normal distance $r$ from the axis of rotation.

We should consider a differential mass $dm$ at a normal distance $r$ from the axis of rotation to compute the moment of inertia of differential mass $dI$ which is $$dI=(\text{Normal distance from axis of rotation})^2\times (\text{Point mass})=r^2dm$$
The moment of inertial $I$ of the system of continuously distributed mass is obtained by integrating $dI$ with proper limits $$I=\int r^2dm$$

For a system consisting of $n$ number of discrete point masses $m_1, m_2, m_3, \ldots m_n$ at normal distances $r_1, r_2, r_3, \ldots r_n$ respectively the moment of inertia $I$ is obtained taking the discrete summation of individual moments of inertia as follows $$I=m_1r_1^2+m_2r_2^2+m_3r_3^2+\ldots +m_nr_n^2=\Sigma_{i=1}^{n}m_ir_i^2 $$