First, a remark that I think is relevant:
understanding the construction $K_{\xi}$ is related to the question of whether a homology class can be represented by a submanifold. This is not always possible;
see e.g. this MO answer.
On the other hand, it is always possible for $n$-cycles for small values of $n$ (if we are considering the homology of a manifold),
which may be one reason why it is hard to visualize non-manifold examples of $K_{\xi}$.
E.g. in the case $n = 2$, we are gluing triangles along their edges, with the rule that exactly two triangles meet along a common edge. This always gives a closed $2$-manifold. In general, unless I am confused, the fact that for $n = 2$ we always get a manifold implies in general that the non-manifold points are in codimension at least $n - 3$ (not just $n - 2$). So to find a counterexample we should take $n = 3$. I will give such a counterexample in a moment, but first let me describe a general approach to thinking about this situation.
When gluing simplices like this, the way to investigate whether the resulting object is a manifold or not is to consider the link of each vertex. Namely, if a bunch of $n$-simplices meet at a vertex $v$, take a transverse slice to each simplex just below the vertex (I am imagining that simplex is sitting on the face opposite to $v$, so that $v$ is at the top of the simplex),
to get an $n-1$-simplex. Now these glue together to make a closed simplicial $n-1$-complex that surrounds $v$; this is the link of $v$. Note that these $n-1$-simplices meet along $n-2$-dimensional faces, and so the link is a $K_{\xi}$-type object, but of one dimension less.
In the case that $n = 2$, we get a bunch of segments being joined at their vertices, and hence a circle. No drama there.
But now suppose that $n = 3$. Then the link is formed by a bunch of triangles gluing together, and we have agreed that this gives a $2$-manifold. But which one?
If $K_{\xi}$ is locally Euclidean at $v$, then this surface would have to be a
$2$-sphere. But a priori it doesn't have to be, and so in this case we can
get a non-manifold example!
In practice, to make an example we should take a cone on a surface that is not
a $2$-sphere, e.g. a cone on a $2$-torus.
Precisely: triangulate a $2$-torus, e.g. by choosing two triangles
$\Delta_1$ and $\Delta_2$, and identifying the three edges of the first with the three edges of the second in the appropriate manner.
Now to form the cone on this, replace $\Delta_i$ by a $3$-simplex
$\tilde{\Delta_i}$. Regard $\Delta_i$ as one of the faces of $\tilde{\Delta}_i$,
and label the other three faces according to the edge along which they
meet $\Delta_i$. Now glue the three faces of $\tilde{\Delta}_1$ other than
$\Delta_1$ with the three faces of $\tilde{\Delta}_2$ other than $\Delta_2$
according to the same gluing scheme we used previously to construct the $2$-torus. What we end up with is a three dimensional simplicial complex whose boundary is equal to $\Delta_1 + \Delta_2$, i.e. a $2$-torus. But it is not a $3$-manifold; rather it is a cone on the $2$-torus.
If we let $\xi$ be the $3$-chain $\tilde{\Delta}_1 + \tilde{\Delta}_2$,
then $K_{\xi}$ is the cone on the $2$-torus, and so is not a $3$-manifold.
Of course, this $\xi$ is not a cycle (so the boundary of $K_{\xi}$ is non-empty), but we could take two such cones and glue
them along their common $2$-torus boundary to get a three-dimensional simplicial complex without boundary, and then take $\xi$ to be the sum of the four $3$-simplices involved; then $\xi$ would be a cycle, but $K_{\xi}$ would not be a manifold; it has two vertices whose links are $2$-tori rather than $2$-spheres.
Concluding remark: In dimension $2$, the only way to get pseudo-manifolds that are not manifolds is to explicitly glue together vertices, as you observed. But in dimension $3$ or higher, there are other examples, e.g. in dimension $3$ we can take cones on positive genus surfaces, as in the preceding construction.
Just to sum up, mostly for my own reference, but I thought others might find it useful. (I am new to the site, so please excuse me if this shouldn't be an answer...)
First some preliminary notions:
For a topological space $X$, an $n$-simplex in $X$ is a continuous map $\Delta^n \to X$ from the standard geometric $n$-simplex $\Delta^n$ into $X$. The maps $d^i: \Delta^{n-1} \to \Delta^{n}$, sends $\Delta^{n-1}$ to the face of $\Delta^n$ sitting opposite the $i$th vertex of $\Delta^n$.
An ordered $n$-simplex is a partially ordered set $n_+ = \{ 0 < 1 < \cdots < n \}$. The $n+1$ elements of $n_+$ is called the vertices of $\sigma$. The subsets of $n_+$ are called the faces of $\sigma$. There are morphisms of simplices $d^i: (n-1)_+ \to n_+$ called coface maps, given by $d^i((n-1)_+) = \{ 0 < 1 < \dots < î < \cdots < n \}$ omitting the $i$th vertex of $n_+$.
Then for the two homologies:
The singular (unreduced) chain complex on a space $X$, is the chain complex
$$\cdots \xrightarrow{\partial_{n+1}} C_n(X) \xrightarrow{\partial_n} C_{n-1}(X) \xrightarrow{\partial_{n-1}} \cdots C_1(X) \xrightarrow{\partial_1} C_0(X) \to 0$$
where $C_n(X)$ is the free abelian group $\mathbb{Z}[S_n(X)]$ generated by the set $S_n(X) = \{ \sigma : \Delta^n \to X \}$ of all $n$-simplices in $X$ (i.e. the set of all continuous maps $\Delta^n \to X$). The boundary maps $\partial_n : C_n(X) \to C_{n-1}(X)$ is given by $\partial_n (\sigma) = \sum_{i=0}^{n}(-1)^i \sigma d^i : \Delta^{n-1} \to \Delta^n \to X$.
The $n$th homology group $H_n(X) = \ker(\partial_n) / \text{im}(\partial_{n+1})$ of this complex is the $n$th singular homology group of $X$.
A simplicial complex $S$ is a set $S = \bigcup_{n=0}^{\infty} S_n$ where $S_n = S(n_+)$ being a set of ordered $n$-simplices, such that a face of any simplex in $S$ is itself a simplex in $S$. The simplicial chain complex
$$\cdots \xrightarrow{\partial_{n+1}} C_n(S) \xrightarrow{\partial_n} C_{n-1}(S) \xrightarrow{\partial_{n-1}} \cdots C_1(S) \xrightarrow{\partial_1} C_0(S) \to 0$$
consists of the free abelian groups $C_n(S) = \mathbb{Z}[S_n]$ generated by the $n$-simplices. The boundary map $\partial_n : C_n(S) \to C_{n-1}(S)$ is given by $\partial_n(\sigma) = \sum_{i=0}^n (-1)^i d_i \sigma$ where $d_i = S(d^i) : S_n \to S_{n-1}$ is the face maps $d_i(\sigma) = \sigma \circ d^i$.
The $n$th homology groups of this complex $H^\Delta_n(S) = \ker(\partial_n) / \text{im}(\partial_{n+1})$ is the $n$th simplicial homology group of $S$.
Lastly we have the realization of $S$, $|S| = \coprod (S_n \times \Delta^n) / \left((d_i \sigma, y) \sim (\sigma, d^iy) \right)$ for all $(\sigma, y) \in S_n \times \Delta^{n-1}$, where $d_i \sigma \times \Delta^{n-1}$ is identified with the $i$'th face of $\sigma \times \Delta^n$.
Then if you want to say something about a specific space $X$, you need to find a simplicial complex $S$, whose realization is homeomorphic to $X$ (i.e. you triangulate $X$ and find the homology groups of the resulting simplicial complex).
NOTE: Feel free to edit any mistakes and clarify where you find it necessary. I'm still not 100% comfortable with it yet..
Best Answer
Basically that is correct, although to call this a "simplicial complex without boundary" is kind of meaningless. There is no such object.
There is a terminology for this object, though, it's called a pseudo-manifold. The idea is that in this quotient simplicial complex, for each point $x$ in the interior of an $n$-simplex or an $n-1$ simplex, there exists an open subset $U$ of the simplicial complex such that $x \in U$ and $U$ is homeomorphic to an open subset of $\mathbb R^n$. So, such points $x$ form an open subset of $X$ which is an oriented $n$-dimensional manifold.
What about the points in faces of dimension $n-2$ or less? Well, there's pretty much nothing to be said, one has no control over their local topology, in particular it can probably be a rather arbitrary $n-2$ dimensional simplicial complex.
Anyway, that's exactly what an $n$-dimensional pseudo-Manifold is: a simplicial complex of dimension $n$ such that the union of the interiors of the $n$ simplices and the $n-1$ simplices is an oriented $n$-dimensional manifold.
But the case $n=2$ is somewhat special, because in this case the pseudo-manifold is actually a manifold: each $0$-simplex also has a neighborhood homeomorphic to an open subset of $\mathbb R^2$. And then yes, in the case $n=2$ you get a 2-dimensional manifold without boundary.