In wiki, the smooth projective model of an affine model is hyperelliptic curve is defined as follows:
https://en.m.wikipedia.org/wiki/Hyperelliptic_curve
To be more precise, the equation defines a quadratic extension of $\mathbb C(x)$ (just the function field of the complex projective line), and it is that function field that is meant. The singular point at infinity can be removed (since this is a curve) by the normalization (integral closure) process. It turns out that after doing this, there is an open cover of the curve by two affine charts: the one already given by
$y^{2}=f(x)$
and another one given by
$w^{2}=v^{{2g+2}}f(1/v)$.
The glueing maps between the two charts are given by
$(x,y)\mapsto (1/x,y/x^{{g+1}})$
and
$(v,w)\mapsto (1/v,w/v^{{g+1}})$
I can see it is smooth by checking at charts, but how to show it is projective i.e proper curve?
The map to P^1 given by taking the x coordinate is finite (check on affine charts). Finite morphisms are proper and therefore the hyperelliptic curve is proper.