Let $X$ be a simplicial set. The partially ordered set $(X^{nd}, \leq)$ of non-degenerate simplices of $X$, with $x\le y$ if $x$ is a face of $y$, was considered by Barratt in a 1956 Princeton preprint. I write $B(X) = N(X^{nd}, \le)$ for its nerve, the Barratt nerve.
It is not quite clear to me if, in your definition of $Face \colon sSet \to Pos$, you only allow the non-degenerate simplices as elements. Let me assume that you make that restriction, so that $Nerve \circ Face(X) = B(X)$.
If $X$ is the simplicial set associated to an (ordered) simplicial complex, then $B(X)$ is
indeed the barycentric subdivision of that simplicial complex. In these cases, there is a homeomorphism $|B(X)| \cong |X|$. However, if $X = \Delta^n/\partial\Delta^n$ for $n\ge1$,
then $X^{nd}$ is isomorphic to $[1] = (0 < 1)$ and its nerve is contractible. So in these cases $|B(X)|$ is not homeomorphic (nor homotopy equivalent) to $|X|$.
From the definition of Kan's normal subdivision $Sd(X)$ as a coend, there is a canonical
natural map $b_X \colon Sd(X) \to B(X)$. It is an isomorphism if and only if $X$ is non-singular, meaning that the representing map $\bar x \colon \Delta^n \to X$ of each non-degenerate simplex of $X$ is a cofibration. Here $n$ is the dimension of $x$.
There exists a homeomorphism $h_X \colon |Sd(X)| \cong |X|$, due to Fritsch and Puppe (1967). However, it is not natural for most maps $X \to Y$ of simplicial sets. For instance, there is no way to fix $h_{\Delta^1}$ and $h_{\Delta^2}$ so that the homeomorphisms are compatible with both of the degeneracy maps $\sigma^j \colon \Delta^2 \to \Delta^1$.
The normal subdivision $Sd(X)$ of any simplicial set is a regular simplicial set, in the sense that each representing map $\bar x \colon \Delta^n \to X$ only makes identifications
along the last ($n$th) face. The geometric realization of a regular simplicial set
is a regular CW complex, i.e., if the closure of each $n$-cell is an $n$-ball and the boundary of the open $n$-cell in its closure is an $(n-1)$-sphere. Any regular CW complex admits a triangulation. See e.g. Fritsch and Piccinini (1990), sections 3.4 and 4.6 for proofs.
Combining the two previous paragraphs, the realization of any simplicial set can be
triangulated by a simplicial complex, but not in a natural way. Not every CW complex can be triangulated (Metzler, 1967), but the realizations of (regular) simplicial sets can.
In work of Waldhausen, Jahren and myself (Spaces of PL Manifolds and Categories of Simple Maps,
Annals of Maths. Studies, to appear in 2013) we consider finite simplicial sets $X$
and prove in Proposition 2.5.8 that there are natural maps
$$
Sd(Sd(X)) \to B(Sd(X)) \to Sd(X) \to X .
$$
Each of these are simple, in the sense that their geometric realizations have contractible
point inverses. In particular, they are simple-homotopy equivalences. The term
$B(Sd(X))$
is the nerve of a partially ordered set, hence an ordered simplicial complex. So there
is, indeed, a natural simple map
$$
i_X \colon B(Sd(X)) \to X
$$
from an ordered simplicial complex to $X$.
You mentioned geometric topology at the outset. If $X$ is a combinatorial manifold,
so that $M = |B(Sd(X))|$ and $N = |X|$ are PL manifolds, then Cohen (1970) proved that
any simple PL map $M \to N$ can be uniformly approximated by a PL homeomorphism (assuming
the manifolds are of dimension at least 5). Hence the natural map $|i_X| \colon |B(Sd(X))| \to |X|$ is a map from an ordered simplicial complex that is arbitrarily
close to a PL homeomorphism, but I have no reason to expect that this
approximation can be made natural.
Best Answer
Here are necessary and sufficient conditions for an abstract, finite simplicial complex $\mathcal{S}$ to be the order complex of some partially ordered set.
(i) $\mathcal{S}$ has no missing faces of cardinality $\geq 3$; and
(ii) The graph given by the edges (=$1$-dimensional simplices) of $\mathcal{S}$ is a comparability.
[Definitions. (a) A missing face of $\mathcal{S}$ is a subset $M$ of its vertices (=$0$-dimensional simplices) such that $M \not \in \mathcal{S}$, but all proper subsets $P\subseteq M$ satisfy $P\in \mathcal{S}$. (b) A graph (=undirected graph with no loops nor multiple edges) is a comparability if its edges can be transitively oriented, meaning that whenever edges $\{p, r_1\}, \{r_1, r_2\},\ldots, \{r_{u−1}, r_u\}, \{r_u, q\}$ are oriented as $(p, r_1), (r_1, r_2),\ldots, (r_{u−1}, r_u), (r_u, q)$, then there exists an edge $\{p, q\}$ oriented as $(p, q)$.]
This characterisation appears with a sketch of proof $-$ which is not hard, anyway $-$ in
As Bayer points out, the result was first observed in
@Rasmus and @Gwyn: The characterisation might perhaps disappoint you if you were expecting something more topological. However, it's easy to prove that no topological characterisation of order complexes is possible, and therefore a combinatorial condition such as the one on comparabilities must be used. For this, first check that the barycentric subdivision of any simplicial complex indeed is an order complex. Next observe that barycentric subdivision of a simplicial complex does not change the homeomorphism type of the underlying polyhedron of the complex. Finally, conclude that for any topological space $T$ that is homeomorphic to a compact polyhedron, there is an order complex whose underlying polyhedron is homeomorphic to $T$.
I hope this helps.