$$\hat p_r^2 = \hat p_r \hat p_r = (-i \hbar)^2\frac1 r \frac d {dr} r\frac1 r \frac d {dr} r=-\hbar^2\frac {d^2} {dr^2}r$$
This probably means:
$$\hat p_r =-i\hbar \frac1 r \frac d {dr} r$$
I'm trying to understand what is happening here. I understand the part $(-i \hbar)^2=-\hbar^2$, but not the rest. I would expect the result to be something like $-\hbar^2 * \frac1 r^2 * \frac{d^2} {dr^2} * r^2 $.
The way it seems to me now, is that multiplying operators isn't the same as multiplying numbers? But instead it means applying the first operator to the result of the later operator? That would mean (here I leave out the constants) applying $\frac1 r \frac{d} {dr} r$ to $\frac1 r \frac{d} {dr} r$ and $r * \frac1 r $ is $1$ and then we get the result.
But then, why is $\nabla * \nabla$ a dot product, considering that multiplication doesn't work the simply anymore? For example, it seems like when multiplying operators, you can't switch around order anymore. So why does the dot product rule still apply? After all, $ \Delta$ yields a number, but $\nabla$ is a vector and still yields a vecor right? Simply applying $\nabla$ twice instead of $\nabla * \nabla = \Delta$ would still yield a vector, right?
$$ \hat{p}_r f = - i \hbar \frac{1}{r} \frac{d}{dr} (r f)\\ \hat{p}_r^2 f = - i \hbar \frac{1}{r} \frac{d}{dr} ( r (\hat{p}_r f))\\ = - \hbar^2 \frac{1}{r} \frac{d}{dr} ( r \frac{1}{r} \frac{d}{dr} (rf))\\ = - \hbar^2 \frac{1}{r} \frac{d}{dr} ( \frac{d}{dr} (rf))\\ = - \hbar^2 \frac{1}{r} \frac{d^2}{dr^2} (rf)\\ $$
Multiplication by $r$ and operating $\frac{d}{dr}$ do not commute so keep the order in mind.
$$ r \frac{d}{dr} f = r \frac{df}{dr}\\ \frac{d}{dr} (rf) = f + r \frac{df}{dr}\\ \frac{d}{dr} ( r f) \neq r \frac{d}{dr} f\\ [r , \frac{d}{dr}] = -1 $$
In this case you aren't bothering with the $\hat{p}_\theta$ or $\hat{p}_\phi$ so no need to make $\nabla$