Suppose $f$ is analytic on an open set $U$ containing the compact set $K$, and $\{r_n\}$ is a sequence of rational functions provided by Runge's theorem (having poles in some prescribed set $A$). For a given finite set $\{x_1,\ldots,x_k\}\subset K$, can we additionally demand that for each $n$,
$r_n(x_1)=f(x_1),r_n(x_2)=f(x_2),\ldots,r_n(x_k)=f(x_k)$?
If yes, can we also require that the sequence match an arbitrarily large set of derivative data. That is, perhaps I want each $r_n$ to match the values of the first $m$ derivatives of $f$ at the same collection of points $\{x_i\}$.
Below I'll be brief with the notation.
Claim: If $E$ is a finite subset of $K$ and $f = 0$ on $E,$ then there are $r_n \to f$ with $r_n = 0$ on $E$ for each $n.$ Proof: Suppose $E$ has just one point $z_1.$ Use Runge to find $r_n \to f.$ Then $r_n-r_n(z_1)$ will do the job, since $r_n(z_1)\to f(z_1) = 0.$
Now assume the claim is true for any $E\subset K$ having $k$ elements. If now $E= \{z_1,\dots ,z_{k+1}\}\subset K$ and $f=0$ on $E,$ then by the induction hypothesis there are $r_n \to f$ with $r_n = 0 $ on $\{z_1,\dots ,z_{k}\}.$ Define
$$s_n(z) = r_n(z) - r_n(z_{k+1})\frac{(z-z_1)\cdots (z-z_k)}{(z_{k+1}-z_1)\cdots (z_{k+1}-z_k)}.$$
Then $s_n = 0$ on $E.$ Because $r_n(z_{k+1})\to f(z_{k+1})=0,$ $s_n \to f.$ This proves the claim.
For the general problem, suppose $E= \{z_1,\dots ,z_{k}\}\subset K.$ Then there is a polynomial $p$ such that $p(z_j) = f(z_j), j = 1, \dots ,k.$ By the claim, there are $r_n \to f-p$ with $r_n = 0$ on $E.$ This implies $r_n+p \to f,$ with $r_n = f$ on $E$ for each $n.$
I haven't thought about interpolating the derivative values.