When creating the splitting field of a polynomial we can without loss of generality assume it has constant coefficient $1$ and splits as $\prod_{k=1}^n(1-z_kT)$. The splitting field is obtained by adjoining $z_k$s and Vieta tells us that $a_k=(-1)^ke_k(z_1,\cdots,z_n)$ where $a_k$ is the $k$th coefficient of the polynomial and $e_k$s are the elementary symmetric polynomials.
Form $R$ as the quotient of $K[Z_1,\cdots,Z_n]$ by the relations $a_k=(-1)^ke_k(Z_1,\cdots,Z_k)$. The ring $R$ will not generally be a field, but every residue field of $R$ (quotient by a maximal ideal) will be a splitting field for $f$ over $K$ (since they will be extensions of $K$ in which $f$ splits generated by these very roots of $f$, the images of $Z_1,\cdots,Z_n$ in the residue field).
Suppose $f(T)\in 1+T K[[T]]$ is an infinite polynomial with constant coefficient $1$. Form $R$ as the quotient of $K[[Z_1,Z_2,\cdots]]$ by the relations $a_k=(-1)^ke_k(Z_1,Z_2,\cdots)$ again. Will all residue fields of $R$ necessarily be isomorphic? What about iso via $Z_i\mapsto Z_{\sigma(i)}$ for some $\sigma\in{\rm Perm}(\Bbb N)$?
Thanks to discussion with Lubin, I see I need to be more explicit about what $K[[Z_1,Z_2,\cdots]]$ is supposed to mean. After toying with some possibilities, I think $\varprojlim K[Z_1,\cdots,Z_n]$ what I want, where the projection maps are evaluation at $0$.