In Mumford's Red Book (Remark III on page 21) he gives the following construction:
Let $k$ be an algebraically closed field, $X=V(xw-yz)\subset k^4$ and $U=D_X(y)\cup D_X(w)$. Define $h\in\Gamma(U,\mathcal{O}_X)$ by setting $h=\frac{x}{y}$ on $D_X(y)$ and $h=\frac{z}{w}$ on $D_X(w)$.
Suppose that $h$ is of the form $\frac{f}{g}$ on $U$. Then let $Z=V(y,w)$, this is a plane in $X$ and $U=X\setminus Z$. We have that $V(g)\cap X\subset Z$ by assumption, and all components of $V(g)\cap X$ have dimension $2$.
I'm struggling to justify the statement in bold. He refers us to Chapter 7, but I can't find a result which explains it. His Corollary 3 there seems to say that if $W$ were a component of $V(h)$ viewed as a closed subset of $X$, then $\text{codim}(W)\leq1$ and so $\dim(W)\geq3$, but this can't be the case since it is contained in a plane.
I know that for irreducible affine algebraic sets $A,B\subset k^n$ we have, for any component $W$ of $A\cap B$, that $\dim(W)\geq\dim(A)+\dim(B)-n$. Then since $\dim(X)=4$, if I could show that $\dim(V(g))=2$ I think we would be done since $\dim(V(g)\cap X)\leq\dim(Z)=2$ because $V(g)\cap X\subset Z$. However I can't see why $V(g)$ needs to be irreducible, or how to show its dimension is $2$ even if it is.
Any help would be much appreciated.