For $f \in K[x_0,\ldots,x_n]$ then, $(\phi_i^{-1})^*(\phi_i^*(f)) | f$
Lets consider the dehomogenisation defined by:
$ \phi_i^*: K[x_0,\ldots,x_n] \rightarrow K[x_1,\ldots,x_n] $, such that, $\phi_i^*(f)(x_1,\ldots,x_n):=f(x_1,\ldots,x_i,1,x_{i+1},\ldots,x_n)$
and the homogenisation defined by:
$(\phi_i^{-1})^*:K[x_1,\ldots,x_n]\rightarrow K[x_0,\ldots,x_n]$, such that, $(\phi_i^{-1})^*(f)(x_0,\ldots,x_n):=x_i^df(\frac{x_0}{x_i},\ldots,\frac{x_{i-1}}{x_i},\frac{x_{i+1}}{x_i},\ldots,\frac{x_n}{x_i})$
Let $f\in K[x_0,\ldots,x_n]$, and $p=(p_o,\ldots,p_n)$, then, $$(\phi_i^*)(f)(x_1,\ldots,x_n)=f(x_1,\ldots,x_i,1,x_{i+1,\ldots,x_n})$$
Let $g=\phi_i^*(f)$, so $$(\phi_i^{-1})^*(g)(p_o,\ldots,p_n)=p_i^d \cdot g\left(\frac{p_o}{p_i},\ldots,\frac{p_{i-1}}{p_i},\frac{p_{i+1}}{p_i},\ldots,\frac{p_n}{p_i}\right)\\=p_i^d\cdot f\left(\frac{p_o}{p_i},\ldots,\frac{p_{i-1}}{p_i},1,\frac{p_{i+1}}{p_i},\ldots,\frac{p_n}{p_i}\right)$$
But if, $f|g$ there exist some $h$ such that $g=hf$, but I do not see it.
\textbf{EDIT}: I'm taking a course of Hyperelliptic Curves, and those are the introductory thopics, my teacher uses ALGEBRAIC CURVES
An Introduction to Algebraic Geometry WILLIAM FULTON, but they do not defined there. So, by @KReiser comment, if I take $f$ to be homogeneous, then they are equal?. Here a scren shot of my virtual class
Because if $d=$ total degree of $f$
$$p_i^d\cdot f\left(\frac{p_0}{p_i},\ldots,\frac{p_{i-1}}{p_i},1,\frac{p_{i+1}}{p_i},\ldots,\frac{p_n}{p_i}\right)=p_i^d\frac{1}{p_i^d}f(p_0,\ldots,p_n)=f(p_0,\ldots,p_n)$$