Deriving the first moment of Collisionless Boltzmann Equation in Spherical Polar Coordinates
After correspondence with the author of those notes, I arrived at the following conclusion.
When we do $\frac{\partial f}{\partial r}$, we require that all other (other than $r$) variables which f is dependent on, are held constant. That is: $\theta, \phi, p_r, p_{\theta}, p_{\phi}$ are constants.
Note that:
$(p_r, p_{\theta}, p_{\phi})=(\dot{r}, r^2 \sin^2 \theta \dot{\phi}, r^2\dot{\theta})$
Keep in mind that:

$p_r$, ie $\dot{r}$ is held constant: it does not mean $r$ is held constant (in fact, we are differentiating with respect to $r$).

$p_{\theta}$, ie $r^2\sin^2\theta\dot{\phi}$ is held constant. This does not mean that either $r$, $\theta$ or $\dot{\phi}$ is held constant.

$p_{\phi}$, ie $r^2\dot{\theta}$ is held constant. We are not claiming this about $r$ or $\dot{\theta}$.


Writing the steps emphasizing this statement above:
$$\int d\vec{p} p_r \frac{\partial f}{\partial r}  p_r \Bigr|_{\theta, \phi, p_r, p_{\theta}, p_{\phi}}$$
Writing out explicitly what we mean by $d\vec{p}$:
$$=\int \mathrm{d}(p_r,p_{\theta},p_{\phi}) p_r \frac{\partial f}{\partial r}  p_r \Bigr|_{\theta, \phi, p_r, p_{\theta}, p_{\phi}}$$
We can move out the partial differentiation because all other term than $\frac{\partial f}{\partial r}$ is held constant: [THIS IS THE MAIN STEP]
$$=\frac{\partial}{\partial r}\int \mathrm{d}(p_r,p_{\theta},p_{\phi}) p_r^2 f \Bigr|_{\theta, \phi, p_r, p_{\theta}, p_{\phi}}$$
Changing from the $p$-nomenclature to the variables they represent:
$$\frac{\partial}{\partial r} \int \mathrm{d}(\dot{r}, r^2\dot{\theta}, r^2\sin^2\theta\dot{\phi})  v_r^2 f \Bigr|_{\theta, \phi, \dot{r}, r^2 \sin^2 \theta \dot{\phi}, r^2\dot{\theta}}$$
ie
$$\frac{\partial}{\partial r} \int \mathrm{d}(v_r, r \sin\theta v_{\phi},  r v_{\theta}) v_r^2 f \Bigr|_{\theta, \phi, \dot{r}, r^2 \sin^2 \theta \dot{\phi}, r^2\dot{\theta}}$$
This step I am unsure about (ie why I don't have a $dr$ and $d\theta$ term, for example), but we say:
$$\mathrm{d}(v_r, r \sin\theta v_{\phi},  r v_{\theta}) = r^2 \sin\theta \mathrm{d}v_r \mathrm{d}v_{\theta} \mathrm{d}v_{\phi}$$
Rewrite integral
$$\frac{\partial}{\partial r} \int r^2 \sin\theta \mathrm{d}v_r \mathrm{d}v_{\theta} \mathrm{d}v_{\phi} v_r^2 f$$
$$=\frac{\partial}{\partial r} r^2 \sin\theta \int \mathrm{d}v_r \mathrm{d}v_{\theta} \mathrm{d}v_{\phi} v_r^2 f$$
$$=\frac{\partial}{\partial r} \left( r^2 \sin\theta \nu \bar{v_r^2} \right)$$
As claimed by the notes.

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