The function $f(z) = \frac{1}{\sqrt{z}\sqrt{1-z}}$ with branch cuts chosen so that $f$ is analytic on $\mathbb C-[0,1]$ has a pole at infinity according to this walkthrough of a branch cut contour integration.
But rather it seems like what I should say is that the one-form $f(z)dz$ has a pole at infinity, since without the $dz$ the function $f(\frac 1w)$ doesn't blowup at $w=0$. Can someone explain what is going on?
I've also read elsewhere that one-forms are automatically coordinate-independent. I'm not sure what exactly is meant by this, but perhaps it is related? That is, we should really talk about poles of objects that are coordinate-independent as opposed to functions which are just one-forms with respect to some implicit choice of coordinates and the differential suppressed -- or something like that. I would appreciate an answer to both questions.
That function doesn't have a pole at infinity. The form does. The reason why you look at the form is because that (the form) is what is inside the integral.
For example. Take the function $g(z)=1$. It doesn't have a pole anywhere. It is constant. But $1\text{d}z=-\frac{1}{w^2}\text{d}w$ has a pole at infinity.
An important observation is that the residue of a function (if you want to define residue of a function) might be non-zero at a regular point, when that point is infinity.
This is why the theory looks nicer when you instead define residues for forms, instead of for functions. At a regular point of a form, the residue is zero.
In the note you linked, they say "the trick is to realize that there is a pole of $f(z)\text{d}z$". There is no trick, or if anything the trick is the use of the change of coordinates $w=1/z$. You don't even need to think about any poles of anyone. You want to compute $\int_C f(z)\text{d}z$. Cauchy's theorem is not applicable to this integral. You change variable to $w=1/z$ such that you can apply Cauchy's theorem. When you change variable you do get a function $1/\sqrt{1-w}$ for which Cauchy's theorem is applicable to the contour integral $\int \frac{1}{\sqrt{1-w}}\frac{\text{d}w}{w}$. You use it, and you are done. No definition of pole necessary.