I suppose you are talking about affine algebraic sets, say $X\subset\mathbb{A}^n$, $Y\subset\mathbb{A}^m$. If $\phi:X\to Y$ is a regular map, then $\phi=(\phi_1,...,\phi_m)$, where $\phi_i\in k[x_1,...,x_n]$. This determines a map $k[Y]=k[x_1,...,x_m]/I(Y)\to k[x_1,...,x_n]/I(X)=k[X]$ via $g\mapsto g(\phi_1,...,\phi_m)$.
Now if there is a $k$-algebra homomorphism $\psi:k[Y]\to k[X]$, we can construct a regular map $\phi:X\to Y$ by setting $\phi=(\psi(x_1),...,\psi(x_m))$.
These mappings are inverse to each other: The image of a regular map $\phi:X\to Y$ is of the form $g\mapsto g(\phi_1,...,\phi_m)$. In particular, $x_i$ is mapped to $\phi_i$ again. So if we use the definition of our second mapping to get back a regular map $X\to Y$, it will be $\phi$ itself again.
Now if we are given $\psi:k[Y]\to k[X]$, we get the regular map $(\psi(x_1),...,\psi(x_m))$, which is then mapped to $g\mapsto g(\psi(x_1),...,\psi(x_m))=\psi(g)$.
You should verify that everything is well-defined here. I hope I made no mistakes, I only know a very similar proof that there is a 1-1 correspondence between morphisms of affine algebraic sets and $k$-algebra homomorphisms of the respective coordinate rings. But I found a little pdf where it's treated with regular maps as well. As I see it, the mentioned 1-1 correspondence ensures that morphisms are basically "the same" as regular (=polynomial) maps between affine algebraic sets. Chapter 2 of this text might be another interesting reference.
If there is anything wrong with the above, please feel free to correct / inform me!