The goal of this post is to describe a partial order on the collection of simplical complexes with $\leqslant n$ unlabeled vertices that is nice in the context of the space $X=\Ran^{\leqslant n}(M)\times \R_{>0}$.
First note that there is a natural order on (abstract) simplicial complexes, given by set inclusion. Interpreting elements of $X$ as simplicial complexes induces a more restrictive order, as new vertices must "split off" from existing ones rather than just be introduced anywhere. Also note that the category usually denoted by $SC$ of simplicial complexes and simplicial maps contains objects with unordered vertices. Here we assume an order on them and consider the action of the symmetric groups to remove the order.
Definition: Let $SC_k$, for some positive integer $k$, be the collection of simplicial complexes with $k$ uniquely labeled vertices. This collection is a poset, with $S\leqslant T$ iff $\sigma\in T$ for every $\sigma\in S$.
The symmetric group on $k$ elements acts on $SC_k$ by permuting the vertices, and taking the image under this action we get $SC_k/S_k$, the collection of simplicial complexes with $k$ unlabeled vertices. This set also has a partial order, with $S\leqslant T$ in $SC_k/S_k$ iff $S'\leqslant T'$ in $SC_k$, for some $S'\in q_k^{-1}(S)$ and $T'\in q_k^{-1}(T)$, where $q_k:SC_k \twoheadrightarrow SC_k/S_k$ is the quotient map.
Definition: For all $i=1,\dots,k$, let $s_{k,i}$ be the $i$th splitting map, which splits the $i$th vertex in two. That is, if the vertices of $S\in SC_k$ are labeled $v_1,\dots,v_k$, then $s_{k,i}$ is defined by
\[\begin{array}{r c l}
s_{k,i}\ :\ SC_k & \to & SC_{k+1}, \\
S & \mapsto & \left\langle S'\cup \{v_i,v_{i+1}\} \cup \displaystyle \bigcup_{\{v_i,w\}\in S'} \{v_{i+1},w\} \right\rangle ,
\end{array}\]where $S'$ is $S$ with $v_j$ relabeled as $v_{j+1}$ for all $j>i$, and $\langle T\rangle$ is the simplicial complex generated by $T$.
By "generated by $T$" we mean generated in the Vietoris-Rips sense, that is, if $\{v_a,v_b\}\in T$ for all $a,b$ in some indexing set $I$, then $\{v_c\ :\ c\in I\}\in \langle T\rangle$. The $i$th splitting map is essentially the $i$th face map used for simplicial sets.
Let $A = \bigcup_{k=1}^n SC_k/S_k$. The splitting maps induce a partial order on $A$, with $S\leqslant T$, for $S\in SC_k/S_k$ and $T\in SC_{k+1}/S_{k+1}$, iff $s_{k,i}(S')\leqslant T'$ in $SC_k$, for some $S'\in q_k^{-1}(S)$, $T'\in q_{k+1}^{-1}(T)$, and $i\in \{1,\dots,k\}$. This generalizes via composition of the splitting maps to any pair $S,T\in A$, and is visually decribed by the diagram below.
Now, let $M$ be a smooth, compact, connected manifold embedded in $\R^N$, and $X=\Ran^{\leqslant n}(M)\times \R_{>0}$. Let $f:X\to A$ be given by $(P,t)\mapsto VR(P,t)$, the Vietoris-Rips complex around the points of $P$ with radius $t$.
Proposition: The map $f:X\to A$ is continuous.
Proof: Let $S\in A$ and $U_S \subseteq A$ be the open set based at $S$. Take any $(P,t)\in f^{-1}(U_S)\subseteq X$, for which we will show that there is an open ball $B\owns (P,t)$ completely within $f^{-1}(U_S)$.
Case 1: $t\neq d(P_i,P_j)$ for all pairs $P_i,P_j\in P$. Then set
\[\epsilon = \min\left\{t, \min_{i<j} |t-d(P_i,P_j)|, \min_{i<j} d(P_i,P_j) \right\}.\]Set $B = B^{\Ran^{\leqslant n}(M)}_{\epsilon/4}(P) \times B^{\R_{>0}}_{\epsilon/4}(t)$, which is an open neighborhood of $(P,t)$ in $X$. It is immediate that $f(P',t')$, for any other $(P',t')\in B$, has all the simplices of $f(P,t)$, as $\epsilon \leqslant |t-d(P_i,P_j)|$ for all $i<j$. If $P_i$ has split in two in $P'$, then for every simplex containing $P_i$ in $f(P,t)$ there are two simplices in $f(P't')$, with either of the points into which $P_i$ split. That is, there may be new simplices in $f(P',t')$, but $f(P',t')$ will be in the image of the splitting maps. Equivalently, $f(P,t)\leqslant f(P',t')$ in $A$, so $B\subseteq f^{-1}(U_S)$.
Case 2: $t= d(P_i,P_j)$ for some pairs $P_i,P_j\in P$. Then set
\[\epsilon = \min\left\{t, \min_{i<j \atop t\neq d(P_i,P_j)} |t-d(P_i,P_j)|,\ \min_{i<j} d(P_i,P_j) \right\},\]and define $B$ as above. We are using the definition of Vietoris-Rips complex for which we add an edge between $P_i$ and $P_j$ whenever $t>d(P_i,P_j)$. Now take any $(P',t')\in B$ such that its image and the image of $(P,t)$ under $f$ are both in $SC_k/S_k$. Then any points $P_i,P_j \in P$ with $d(P_i,P_j)=t$ that have moved around to get to $P'$, an edge will possibly be added, but never removed, in the image of $f$ (when comparing with the image of $(P,t)$). This means that we have $f(P,t)\leqslant f(P',t')$ in $SC_k/S_k$, so certainly $f(P,t)\leqslant f(P',t')$ in $A$. The same argument as in the first case holds if points of $P$ split. Hence $B\subseteq f^{-1}(U_S)$ in this case as well. $\square$
This proposition shows in particular that $X$ is poset-stratified by $A$
First note that there is a natural order on (abstract) simplicial complexes, given by set inclusion. Interpreting elements of $X$ as simplicial complexes induces a more restrictive order, as new vertices must "split off" from existing ones rather than just be introduced anywhere. Also note that the category usually denoted by $SC$ of simplicial complexes and simplicial maps contains objects with unordered vertices. Here we assume an order on them and consider the action of the symmetric groups to remove the order.
Definition: Let $SC_k$, for some positive integer $k$, be the collection of simplicial complexes with $k$ uniquely labeled vertices. This collection is a poset, with $S\leqslant T$ iff $\sigma\in T$ for every $\sigma\in S$.
The symmetric group on $k$ elements acts on $SC_k$ by permuting the vertices, and taking the image under this action we get $SC_k/S_k$, the collection of simplicial complexes with $k$ unlabeled vertices. This set also has a partial order, with $S\leqslant T$ in $SC_k/S_k$ iff $S'\leqslant T'$ in $SC_k$, for some $S'\in q_k^{-1}(S)$ and $T'\in q_k^{-1}(T)$, where $q_k:SC_k \twoheadrightarrow SC_k/S_k$ is the quotient map.
Definition: For all $i=1,\dots,k$, let $s_{k,i}$ be the $i$th splitting map, which splits the $i$th vertex in two. That is, if the vertices of $S\in SC_k$ are labeled $v_1,\dots,v_k$, then $s_{k,i}$ is defined by
\[\begin{array}{r c l}
s_{k,i}\ :\ SC_k & \to & SC_{k+1}, \\
S & \mapsto & \left\langle S'\cup \{v_i,v_{i+1}\} \cup \displaystyle \bigcup_{\{v_i,w\}\in S'} \{v_{i+1},w\} \right\rangle ,
\end{array}\]where $S'$ is $S$ with $v_j$ relabeled as $v_{j+1}$ for all $j>i$, and $\langle T\rangle$ is the simplicial complex generated by $T$.
By "generated by $T$" we mean generated in the Vietoris-Rips sense, that is, if $\{v_a,v_b\}\in T$ for all $a,b$ in some indexing set $I$, then $\{v_c\ :\ c\in I\}\in \langle T\rangle$. The $i$th splitting map is essentially the $i$th face map used for simplicial sets.
Let $A = \bigcup_{k=1}^n SC_k/S_k$. The splitting maps induce a partial order on $A$, with $S\leqslant T$, for $S\in SC_k/S_k$ and $T\in SC_{k+1}/S_{k+1}$, iff $s_{k,i}(S')\leqslant T'$ in $SC_k$, for some $S'\in q_k^{-1}(S)$, $T'\in q_{k+1}^{-1}(T)$, and $i\in \{1,\dots,k\}$. This generalizes via composition of the splitting maps to any pair $S,T\in A$, and is visually decribed by the diagram below.
Now, let $M$ be a smooth, compact, connected manifold embedded in $\R^N$, and $X=\Ran^{\leqslant n}(M)\times \R_{>0}$. Let $f:X\to A$ be given by $(P,t)\mapsto VR(P,t)$, the Vietoris-Rips complex around the points of $P$ with radius $t$.
Proposition: The map $f:X\to A$ is continuous.
Proof: Let $S\in A$ and $U_S \subseteq A$ be the open set based at $S$. Take any $(P,t)\in f^{-1}(U_S)\subseteq X$, for which we will show that there is an open ball $B\owns (P,t)$ completely within $f^{-1}(U_S)$.
Case 1: $t\neq d(P_i,P_j)$ for all pairs $P_i,P_j\in P$. Then set
\[\epsilon = \min\left\{t, \min_{i<j} |t-d(P_i,P_j)|, \min_{i<j} d(P_i,P_j) \right\}.\]Set $B = B^{\Ran^{\leqslant n}(M)}_{\epsilon/4}(P) \times B^{\R_{>0}}_{\epsilon/4}(t)$, which is an open neighborhood of $(P,t)$ in $X$. It is immediate that $f(P',t')$, for any other $(P',t')\in B$, has all the simplices of $f(P,t)$, as $\epsilon \leqslant |t-d(P_i,P_j)|$ for all $i<j$. If $P_i$ has split in two in $P'$, then for every simplex containing $P_i$ in $f(P,t)$ there are two simplices in $f(P't')$, with either of the points into which $P_i$ split. That is, there may be new simplices in $f(P',t')$, but $f(P',t')$ will be in the image of the splitting maps. Equivalently, $f(P,t)\leqslant f(P',t')$ in $A$, so $B\subseteq f^{-1}(U_S)$.
Case 2: $t= d(P_i,P_j)$ for some pairs $P_i,P_j\in P$. Then set
\[\epsilon = \min\left\{t, \min_{i<j \atop t\neq d(P_i,P_j)} |t-d(P_i,P_j)|,\ \min_{i<j} d(P_i,P_j) \right\},\]and define $B$ as above. We are using the definition of Vietoris-Rips complex for which we add an edge between $P_i$ and $P_j$ whenever $t>d(P_i,P_j)$. Now take any $(P',t')\in B$ such that its image and the image of $(P,t)$ under $f$ are both in $SC_k/S_k$. Then any points $P_i,P_j \in P$ with $d(P_i,P_j)=t$ that have moved around to get to $P'$, an edge will possibly be added, but never removed, in the image of $f$ (when comparing with the image of $(P,t)$). This means that we have $f(P,t)\leqslant f(P',t')$ in $SC_k/S_k$, so certainly $f(P,t)\leqslant f(P',t')$ in $A$. The same argument as in the first case holds if points of $P$ split. Hence $B\subseteq f^{-1}(U_S)$ in this case as well. $\square$
This proposition shows in particular that $X$ is poset-stratified by $A$