To have a well defined homogeneous polynomial $f:\mathbb{P}^n\rightarrow \mathbb{P}^n$, we require that it is homogeneous so that $f(x)=f(\lambda x)=\lambda^{\deg f} f(x)=f(x)$ (so that it is well defined on projective space).
In various texts (e.g. Sutherland's lecture notes on Elliptic Curves, Harris Geometry of Algebraic Curves) a rational function $g:\mathbb{P}^n\rightarrow \mathbb{P}^n$ is defined to be the ratio of homogeneous polynomials $f$ of the same degree. Intuitively I assume this is to make sure $g$ is well defined.
However, let $g=f/h,$ where $f,g$ are homogeneous polynomials on projective space of some arbitrary degrees $F,G$. Then, $g(x)=g(\lambda x)=f(\lambda x)/h(\lambda x)=\lambda^{F-G}f(x)/h(x)=g(x)$. This does not require that $F=G$.
Question: Why then is it required that $f$ and $h$ have the same degree for $g=f/h$? Am I wrong in my analysis somewhere, or is there a different reason for the definition, or am I wrong in interpretation of the definition?
Edit: or do we consider $g$ as a function to affine space?
First of all, a ratio of homogeneous polynomials of the same degree defines a function $\mathbb P^n \to \mathbb P^1$, not $\mathbb P^n \to \mathbb P^n$. For the latter, the usual description is as an $(n+1)$-tuple of coordinate maps, each of which is a homogeneous polynomial of the same degree, not all vanishing at any common point.
For example, consider $\mathbb P^1$ with projective coordinates $[x : y]$. If $f, g \in k[x, y]$ are homogeneous of degrees $F, G$ and $h := [f : g]$, then we have: \begin{align*} h([\lambda x : \lambda y]) & = [f(\lambda x, \lambda y) : g(\lambda x, \lambda y)] \\ & = [\lambda^F f(x, y) : \lambda^G g(x, y)], \end{align*} which agrees with $[f(x, y) : g(x, y)]$ if and only if $F = G$.
I think your mistake is in passing between affine and projective coordinates. If $[f : g]$ is a map to $\mathbb P^1$ as above, we can view it in affine coordinates as a ratio $f/g$ of homogeneous polynomials, where the zero locus of $g$ is sent to the point at infinity. But affine coordinates are not defined up to scalar multiplication, so $\lambda^{F-G} f(x)/g(x)$ is not the same thing as $f(x)/g(x)$.