Let $k$ be an algebraically closed field. Take some homogenous irreducible polynomial $F(X_0,X_1,X_2)$ and set, $$ X = \{ (a:b:c) \in \mathbb{P}^2_k ~ | ~ F(a,b,c) = 0 \} $$ i.e. $X$ is a projective plane curve.
Denote by $K=k(X)$ to be the function field of $X$.
Since $K$ is a function field of dimension $1$ we can choose a rational function $f\in K$ which is transcendental over $k$ and $K$ is a finite extension, so pick some $g\in K$ such that $K = k(f)(g)$ (this is always possible even if $k$ has positive characteristic).
Fix some surjective valuation $v:K^{\times}\to \mathbb{Z}$ (which is trivial on $k^{\times}$). Suppose $f,g\in \mathscr{O}_v$ i.e. $v(f),v(g)\geq 0$. Then we can view $f,g$ are being defined at $v$, so $f(v),g(v)\in k$.
Construct the point $p = (f(v):g(v):1)\in \mathbb{P}^2$.
How does one show that $p\in X$ and that $\mathscr{O}_p$ is contained in $\mathscr{O}_v$.
(To clarify, $\mathscr{O}_p$ are rational functions defined at $p$, and $\mathscr{O}_v$ are rational functions defined at $v$. The distinction is that $p$ is a point in the projective plane while $v$ is a "point" on the abstract curve of valuations.)
Your "surjective valuation $K^\times \to k$" doesn't make much sense, did you mean $K\to \Bbb{Z}\cup\infty$. What do you mean with $f(v)$, and what is $p$ and $\mathscr{O}_p$.
Note that $K=k(x)[y]/(F(x,y,1))$. Did you mean $f=x,g=y$, in which case $p=(a:b:1)$ is a point of your curve and $\mathscr{O}_p$ makes sense: the localization of $R=k[x,y]/(F(x,y,1))$ at the maximal ideal $(x-a,y-b)$. Note that I meant that $v$ is any non-trivial valuation on $K$ which is trivial on $k$ such that $v(x)\ge 0,v(y)\ge 0$ and $a,b\in k$ are the images of $x,y$ under the homomorphism $\mathscr{O}_v\to \mathscr{O}_v /\mathfrak{m}_v\cong k$.
$v(x)\ge 0,v(y)\ge 0$ gives that $R\subset \mathscr{O}_v$.
$x-a,y-b\in \mathfrak{m}_v$ so $(x-a,y-b)\subset \mathfrak{m}_v\cap R$, and since $(x-a,y-b)$ is a maximal ideal, it means that $$(x-a,y-b)=\mathfrak{m}_v\cap R$$ which makes it clear that any $h\in R,\not\in (x-a,y-b)$ is not in $\mathfrak{m}_v$ ie. $h^{-1}\in \mathscr{O}_v \implies \mathscr{O}_p\subset \mathscr{O}_v$.