The above image is taken from the wiki page on polar coordinates.
$(1).r = \sqrt{x^2+y^2} \\ (2).x = r\cos(\theta) \\ (3).y = r\sin(\theta) \\ (4).\theta = \tan^{-1}(\frac{r\sin(\theta)}{r\cos(\theta)})$
and the dot product is $\vec{a} \cdot \vec{b} = |a||b|\cos(\alpha)$.
I want to do in vector calculus: $\textbf{r} \cdot \nabla f(r)$. The function is evenly distributed with the highest values at the center and and if I were to contour it, the contours would be perfectly circular; thus the Gradient vector points normal inwards from the circumference at all points on the boundary and the radius vector is also always normal to the circumference and therefore antiparallel to the Gradient vector so the angle $\alpha = 180$ and we would get $-|\nabla f(x,y)||r|$.
So my question is what would $|r|$ look like if I were to write it out? $r$ in equation (1) is already a magnitude just by looking at the equation? Would it look like $||\vec{r}|| = |\sqrt{x^2+y^2}|$? Or is equation (1) a vector and not a magnitude? Thank you!
The $r$ in the formulas is the coordinate, which is a scalar quantity. By your description of $f$, it is purely a function of this coordinate $r$ (and independent of $\theta$).
You will have to distinguish between this coordinate, and a general position vector $\mathbf{r}$, which I'm guessing is what you want in your dot product. It is true that the magnitude of this position vector is the coordinate, that is, $||\mathbf{r}|| = r$.
In fact, you could write $ \mathbf{r} = r \mathbf{e}_r $ where $\mathbf{e}_r$ is the unit vector in the radial direction. For this reason, the dot product will always be $$ \mathbf{r}\cdot \nabla f(r,\theta) = r\frac{\partial f}{\partial r} $$ even if $f$ did depend on $\theta$. In your description of $f$, $\frac{\partial f}{\partial r}$ is always negative, hence your result.