I'm trying to understand some stuff regarding differentials on a hyperelliptic curveof genus $g$. I'm working with an odd model, $C: \;y^2=f(x)$ over $\mathbb{A}^2_F$, where $F$ is a characteristic $0$ field. I'm looking at the de Rham cohomology $H^1_{DR}(C)$. I get why the basis of this cohomology is given by
$$\omega_0 = \dfrac{dx}{y}, \omega_1= x\omega_0,\omega_2= x^2\omega_0,...,\omega_{2g-1}= x^{2g-1}\omega_0 $$
where if projective co-ordiantes are $[X:Y:Z]$ then $x = \dfrac{X}{Z}$, $y= \dfrac{Y}{Z}$. What I am trying to get my head around is the following:
At infinity, $x$ has a pole of order $2$ and $y$ has a pole of order $2g+1$. This then means that $\omega_0, \omega_1,...,\omega_{g-1}$ are all holomorphic at infinity, and can be extended to regular differentials on the complete hyperelliptic curve. However, $\omega_g,...,\omega_{2g-1}$ are not holomorphic and have in fact got poles of orders $-2,-4,..,-2g$ at infinity. Therefore, they do not extend to regular differentials at on the complete elliptic curve. I have been trying to understand this, but everywhere I go it seems like it is simply asserted.
I've tried to understand the corresponding statement for elliptic curves - i.e. that $\omega=\dfrac{dx}{y}$ and $xw$ are regular and of the second kind respectively, but again it doesn't seem clear to me. If we take, as suggested in other places, the local co-ordainte $t = \dfrac{x^{g}}{y}$ then it would come out pretty quickly, but really this to me is using what we want to show to show exactly that.
Needless to say, I'm a bit confused and would appreciate a nudge in the right direction. I guess it has something to do with the fact that we'd want to consider the class of the divisors $(\dfrac{X}{Z}),(\dfrac{Y}{Z})$ coming out of $K(C)$ and then consider the valuation $v_{\infty}$ at these classes, but I'm struggling a bit.