Suppose that $X = \mathbb{A}^2$. Let $Y$ be the blow up of $X$ at the maximal ideal $(x,y)$ and let $W$ be the blow up of $Y$ at a point on the exceptional divisor of $Y$ over $X$. Of course, the composition $f: W \rightarrow X$ is birational and an isomorphism away from the origin. The fiber of $f$ over the origin is the union of two $\mathbb{P}^1$'s meeting at a single point, but the total space $W$ is non-singular. The map $f$ is identified with the blowup of $X$ along some closed subscheme $Z$ of $X$ supported only at the origin. I believe an example of an ideal defining such a $Z$ is $(x^3, xy, y^2)$.
By taking the composition of blowups along smooth centers, there is some ideal sheaf on the base giving the composition in "one step". In theory, you can compute this ideal by tracing through the proof that every birational morphisms is a blow up - but in practice I think this is usually difficult.
ADDED: here is a proof of the statement you need (namely the square free monomial ideal $I$ is a intersection of primes generated by subsets of parameters) without using the modularity property. We will use induction on $N=$ the total numbers of times the parameters appear in the generators of $I$. For example if $I=(xy, xz)$ then $N=4$. The statement is obvious if $N=1$.
Suppose $I$ has a generator (say $f_1$) which involves at least $2$ parameters. Pick one of these parameters, say $x$ and WLOG, we can assume $I=(f_1,\cdots, f_n, g_1,\cdots,g_l) $ such that $x|f_i$ for each $i$ but $x$ does not divide any of the $g_j$s. Let $F_i=f_i/x$. We claim that:
$$ I = (I,x) \cap (I,F_1)$$
If the claim is true, we are done by applying the induction hypothesis to $(I,x)$ and $(I,F_1)$. One containment is obvious, for the other one we need to show if $xu \in (I,F_1)$ then $xu\in I$.
Write $$xu = f_2x_2 + \cdots f_nx_n + \sum g_jy_j + F_1x_1$$ which implies
$$x(u- F_2x_2 +\cdots F_nx_n) \in (g_1,\cdots, g_l, F_1) = I' $$
$I'$ has minimal generators which do not contain $x$. By induction, $I'$ is an intersection of primes generated by other parameters, so $x$ is a NZD on $R/I'$. So $(u- F_2x_2 +\cdots F_nx_n) \in I'$, and therefore $xu \in I$, as desired.
REMARK: note that for this proof to work, you only need that all subsets of the sequence (not necessarily parameters) generate prime ideals. I guess it fits with your other question.
So from the comments I will take your question as proving $J\cap (K+L) = J\cap K + J \cap L$ for parameter ideals (by which I mean ideals generated by subsets of a fixed regular s.o.p).
It will suffice to understand $I\cap J$ for two such ideals. To be precise, let $g(I)$ be the set of s.o.p generators of $I$. Let $P$ be the ideal generated by the intersection of $g(I),g(J)$, and $I', J'$ generated by $g(I)-g(P), g(J)-g(P)$. Then we need to show:
$$I \cap J = P + I'J' $$
Since $R/P$ is still regular we can kill $P$ and assume that $g(I), g(J)$ are disjoint, and we have to prove $I \cap J = IJ$. This should be an easy exercise, but a slick and very general way is invoking Tor (which shows that this is even true for $I,J$ generated by parts of a fixed regular sequence).
Best Answer
The quoted result relies on the following elementary characterization of local regular rings: