Based on the comments under the question, it seems that the real question is as follows:

Let $P$ be a finite poset whose maximal chains have length $\leq n+1$ for some strictly positive $n \in \mathbb{N}$. Assume the existence of a Morse matching which only admits critical cells in dimensions $n$ and $0$. Is it true that the order complex $\Delta(P)$ of our original poset is homotopy-equivalent to a disjoint union of wedges-of-spheres?

The answer is yes, and in fact much more can be concluded.

Let $c_1,\ldots,c_k$ be the critical cells of dimension $n$ and $d_1,\ldots,d_\ell$ be the critical cells of dimension $0$ (one can assume $k=\ell$ if one wishes, but it is completely unnecessary). A *gradient path* from $c_i$ to $d_j$ is an alternating sequence of elements in $P$ given by

$$ c_i > e_1 < \mu(e_1) > e_2 < \mu(e_2) > \cdots > e_p < \mu(e_p) > d_j$$

where $\mu$ denotes our acyclic matching. It follows by acyclicity of $\mu$ that no such path can be a loop. Note also that there can be no paths between two $c$-cells or two $d$-cells by dimension considerations alone. Associate to each critical cell $c_i$ its "unstable manifold" $W_i^-$ consisting of cells on all paths starting from $c_i$, and associate to each critical cell $d_j$ its "stable manifold" $W_j^+$ consisting of cells on all paths ending at $d_j$. It follows from Forman's first discrete Morse lemma that each $W_i^-$ simple homotopy collapses to $c_i$ and $W_j^+$ simple homtopy collapses to $d_j$, so in fact each stable and unstable manifold in sight is contractible.

It is well-known that the union of two contractible simplicial complexes is homotopy-equivalent to the suspension of their intersection. Applying this rule to each critical cell pair $(d_j,c_i)$ tells you not only that you have a disjoint union of wedges of $n$-spheres, but also which spheres are wedged together along which points.