In what follows, we detail the 1-D ($n=1$), 2-D ($n=2$) cases.
Focus on the 1-D case ($n=1$)
Let us first define the column and row data as:
$$
\left\{
\begin{array}{rcl}
P_c^{(1)}&:=&\left\{\left(\lambda_1(j_1);w(j_1)\right), ~j_1=1,\ldots,{k_1}\right\} \\
P_r^{(1)}&:=&\left\{\left(\mu_1(i_1);v(i_1)\right), ~i_1=1,\ldots,{q_1}\right\}
\end{array}
\right. .
$$
It follows the 1-D Loewner matrix:
$$
\mathbb{L}_1\in\mathbb{C}^{q_1\times k_1}
$$
where each entry reads,
$$
(\mathbb{L}_1)_{i_1,j_1} = \dfrac{v(i_1)-w(j_1)}{\mu_{1}(i_1)-\lambda_{1}(j_1)}.
$$
The null-space of the Loewner matrix
$$
\mathbf{c}_1 = \mathcal{N}(\mathbb{L}_1)
$$
where
$$
\mathbf{c}_1=
\left[
\begin{array}{c}
c(1)\\ c(2) \\ \vdots \\ c(k_1) \\
\end{array}
\right]\in \mathbb{C}^{k_1}
$$
provides the barycentric weigths of the Lagrange form
$$
g(x_1) =
\dfrac{\sum_{j_1=1}^{k_1} \dfrac{c(j_1)w(j_1)}{x_1-\lambda_{1}(j_1)}}{\sum_{j_1=1}^{k_1} \dfrac{c(j_1)}{x_1-\lambda_{1}(j_1)}},
$$
which interpolates the data and recovers the underlying 1-D function or 1-D data (under mild conditions).
Focus on the 2-D case ($n=2$)
Let us first define the column and row data as:
$$
\left\{
\begin{array}{rcl}
P_c^{(2)}&:=&\left\{(\lambda_{1}(j_1),\lambda_{2}(j_2);w(j_1,j_2)), ~j_1=1,\ldots,k_1 \quad j_2=1,\ldots,k_2\right\} \\
P_r^{(2)}&:=&\left\{(\mu_{1}(i_1),\mu_{2}(i_2);v(i_1,i_2)),~i_1=1,\ldots,q_1 \quad i_2=1,\ldots,q_2 \right\}
\end{array}
\right. .
$$
It follows the 2-D Loewner matrix:
$$
\mathbb{L}_2\in\mathbb{C}^{q_1q_2\times k_1k_2}
$$
where each entry reads,
$$
\ell_{j_1,j_2}^{i_1,i_2} = \dfrac{v_{i_1,i_2}-w_{j_1,j_2}}{(\mu_{1}(i_1)-\lambda_{1}(j_1))(\mu_{2}(i_2)-\lambda_{2}(j_2))}.
$$
The null-space of the Loewner matrix
$$
\mathbf{c}_2 = \mathcal{N}(\mathbb{L}_2)
$$
$$
\mathbf{c}_2=
\left[
\begin{array}{c}
c(1,1)\\ \vdots \\ c(1,k_2) \\ \hline \vdots \\ \hline
c(k_1,1)\\ \vdots \\ c(k_1,k_2) \\
\end{array}
\right]\in \mathbb{C}^{k_1k_2}
$$
provides the barycentric weigths of the Lagrange form
$$
g(x_{1},x_{2}) =
\dfrac{\sum_{j_1=1}^{k_1}\sum_{j_2=1}^{k_2} \dfrac{c(j_1,j_2)w(j_1,j_2)}{(x_{1}-\lambda_{1}(j_1))(x_{2}-\lambda_{2}(j_2))}}{\sum_{j_1=1}^{k_1}\sum_{j_2=1}^{k_2} \dfrac{c(j_1,j_2)}{(x_{1}-\lambda_{1}(j_1))(x_{2}-\lambda_{2}(j_2))}}
$$