What you describe is similar to something well-known that does satisfy your condition (ie, has the same homotopy as $X$ in dimensions less than $n-k$). I'm speaking of the dual complex to the dual $(n-k)$-skeleton of your simplicial complex. Let me call this $X^{\ast (n-k)\ast}$ (which is short for $((X^\ast)^{(n-k)})^\ast$). This is not a simplicial complex in general, only a cone complex. It has one vertex for each $k$-simplex; vertices corresponding to those $k$-simplices that belong to the same $(k+1)$-simplex are connected by an "edge" (that is, a cone over those vertices - notice that there can be more than two of them); and so on. You can find the discussion of dual complexes in standard texts on PL topology, but cone complexes as such are not clearly exposed there; see references on cone complexes in my answer here.

What you describe is similar to something well-known that does satisfy your condition (ie, has the same homotopy as $X$ in dimensions less than $n-k$). I'm speaking of the dual complex to the dual $(n-k)$-skeleton of your simplicial complex. Let me call this $X^{\ast (n-k)\ast}$ (which is short for $((X^\ast)^{(n-k)})^\ast$). This is not a simplicial complex in general, only a cone complex. It has one vertex for each $k$-simplex; vertices corresponding to those $k$-simplices that belong to the same $(k+1)$-simplex are connected by an "edge" (that is, a cone over those vertices - notice that there can be more than two of them); and so on. You can find the discussion of dual complexes in standard texts on PL topology, but cone complexes as such are not clearly exposed there; see references on cone complexes in my answer here.