I want to start with a trivial example of computation of the stalk of $\mathbb{P}^2$ in some point, for example $p=|1:0:0|$, $\mathcal{O}_{\mathbb{P}^2, p}$.
By definition, the stalk is the (local) ring of the rational functions $\frac{f}{g}$ of $\mathbb{P}^2$ such that $g(p)\neq 0$, $deg(f)=deg(g)$. In our case $p\not \in Z(x_0)$ and so it makes sense to consider $\frac{f}{x_0^n}$ where $n$ is the degree of the homogenous polynomial $f$. Then an element will be $\frac{f}{g}=\frac{\frac{f}{x_0^n}}{\frac{g}{x_0^n}}\in \mathbb{C}[\frac{x_1}{x_0}, \frac{x_2}{x_0}]_{(\frac{x_1}{x_0}, \frac{x_2}{x_0})}$. This means
$\mathcal{O}_{\mathbb{P}^2, p}\cong \mathbb{C}[\frac{x_1}{x_0}, \frac{x_2}{x_0}]_{(\frac{x_1}{x_0}, \frac{x_2}{x_0})}$
that is what we could expect us because $\mathcal{O}_{\mathbb{P}^2, p}\cong \mathcal{O}_{U, p}$ for each open subset of $U$ of $\mathbb{P}^2$.
If I have an irreducible subvariety $X$ of $\mathbb{P}^n$ defined by the equations $I(X)$ in $\mathbb{C}[x_0,\dots , x_n]$, then it is clear that $\mathcal{O}_{X, p}$ is the (local) ring of the rational functions $\frac{f+I(X)}{g+I(X)}$ where $g+I(X)(p)\neq 0$ and $deg(f)=deg(g)$. If we consider $p\not \in Z(x_0)$, then we get $\frac{f}{g}=\frac{\frac{f+I(X)}{x_0^n+I(X)}}{\frac{g+I(X)}{x_0^n+I(X)}}\in \mathbb{C}[\frac{x_1}{x_0}, \frac{x_2}{x_0}, \dots \frac{x_n}{x_0}]/I(X)_{(\frac{x_1}{x_0}+I(X), \cdots , \frac{x_n}{x_0}+I(X))}$. Thus I suppose that
$\mathcal{O}_{X, p}\cong \mathbb{C}[\frac{x_1}{x_0}, \frac{x_2}{x_0}, \dots \frac{x_n}{x_0}]/I(X)_{(\frac{x_1}{x_0}+I(X), \cdots , \frac{x_n}{x_0}+I(X))}$
The next step is to generalize all the argument:
Given a completely arbitrary graded ring $R$ (in particular not supposed to be a domain) and its associated projective scheme $X=\text{Proj}R$, the stalk of the structure sheaf $\mathcal O_X$ at $\mathfrak p\in X$ is $$\mathcal O_{X,\mathfrak p}=R_{(\mathfrak p)}$$ Here $R_{(\mathfrak p)}\subset R_{\mathfrak p}$ is the subring consisting of fractions $\frac r\pi$ where $r\in R, \pi\in R\setminus \frak p$ are homogeneous elements of the same positive degree.
At this point I can express my problem, that is to compute the local ring of the following subvariety $$X:=Z(rk (\begin{pmatrix} z_1^n & \dots & z_k^n \\ l_1 & \dots & l_k \end{pmatrix}=1)\subseteq \mathbb{P}^2\times \mathbb{P^{k-1}} $$
where $l_j$ is the linear form of a line on $\mathbb{P}^2$ and $n$ is a natural number. We fix a point $q=(p, a)$ such that $p$ does not belong in the line $l_1=b_0x_0+b_1x_1+b_2x_2$, for example.
The stalk in this case would be the following
$\mathcal{O}_{X,p}\cong (\mathbb{C}[x_0, x_1, x_2]_{rk \begin{pmatrix} x_0 & x_1 & x_2 \\ p_0 & p_1 & p_2 \end{pmatrix}=1}\otimes \mathbb{C}[z_1, \dots , z_k]_{rk \begin{pmatrix} z_1 & \dots & z_k \\ a_1 & \dots & a_k \end{pmatrix}=1})/(rk \begin{pmatrix} z_1^n & \dots & z_k^n \\ l_1 & \dots & l_k \end{pmatrix}=1)$
but now we can observe $p\not \in Z(l_1)$ and so elements of the type $\frac{f}{g}$ of the local ring $\mathbb{C}[x_0, x_1, x_2]_{rk \begin{pmatrix} x_0 & x_1 & x_2 \\ p_0 & p_1 & p_2 \end{pmatrix}=1}$
can be written as $\frac{\frac{f}{l_1^d}}{\frac{g}{l_1}^d}$ and this permit us to say
$\mathbb{C}[x_0, x_1, x_2]_{rk \begin{pmatrix} x_0 & x_1 & x_2 \\ p_0 & p_1 & p_2 \end{pmatrix}=1}\cong (\mathbb{C}[\frac{x_0}{l_1}, \frac{x_1}{l_1}, \frac{x_2}{l_1}]/(b_0\frac{x_0}{l_1}+b_1\frac{x_1}{l_1}+b_2\frac{x_2}{l_1}-1)_{rk \begin{pmatrix} \frac{x_0}{l_1} & \frac{x_1}{l_1} & \frac{x_2}{l_1} \\ \frac{p_0}{l_1(p)} & \frac{p_1}{l_1(p)} & \frac{p_2}{l_1(p)} \end{pmatrix}=1}\cong \mathbb{C}[\frac{x_1}{l_1}, \frac{x_2}{l_1}]_{( \frac{x_1}{l_1}- \frac{p_1}{l_1(p)} , \frac{x_2}{l_1} -\frac{p_2}{l_2(p)}) }$
On the other side, the relations $a_i^n(q)=a_1(q)^n\frac{l_i(p)}{l_1(p)}$ tells us $a_1(q)$ must be different from zero and so we get
$\mathbb{C}[z_1, \dots , z_k]_{rk \begin{pmatrix} z_1 & \dots & z_k \\ a_1 & \dots & a_k \end{pmatrix}=1}\cong \mathbb{C}[\frac{z_2}{z_1}, \dots , \frac{z_k}{z_1}]_{(\frac{z_2}{z_1}-\frac{a_2}{a_1}, \cdots , \frac{z_k}{z_1}-\frac{a_k}{a_1})}$
At this point we can use the relations $(\frac{z_i}{z_1})^n=\frac{l_i}{l_1}$ to built an isomorphism that maps the generators $\frac{z_i}{z_1}\to (\frac{l_i}{l_1})^\frac{1}{n}$.
Thus we get
$\mathcal{O}_{X,p}\cong \mathbb{C}[\frac{x_1}{l_1}, \frac{x_2}{l_1}]_{( \frac{x_1}{l_1}- \frac{p_1}{l_1(p)} , \frac{x_2}{l_1} -\frac{p_2}{l_2(p)}) }[(\frac{l_2}{l_1})^\frac{1}{n}, \cdots , (\frac{l_k}{l_1})^\frac{1}{n}]_{((\frac{l_2}{l_1})^\frac{1}{n}-\frac{a_2}{a_1}, \cdots , (\frac{l_k}{l_1})^\frac{1}{n}-\frac{a_k}{a_1})}$
Is it correct? I have some doubt.