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.
The proof that $\Sigma$ is a sphere with $m \ge 0$ handles and $1$ hole is an application of the classification theorem for surfaces with boundary: for every compact, connected, oriented surface $\Sigma$ there exists $m \ge 0$ and $k \ge 0$ such that $\Sigma$ is a sphere with $m$ handles and $k$ holes, where $k$ equals the number of components of the boundary of $\Sigma$; and your surface $\Sigma$ is a compact, connected surface whose boundary is connected, corresponding to the loop $f$.
The reason that $m$ corresponds to a product of $m$ commutators is because if $\Sigma$ is a sphere with $m$ handles and $1$ hole then $\pi_1(\Sigma)$ is a free group having a free basis consisting of $2m$ elements $a_1,b_1,...,a_m,b_m$, and the boundary of the surface, which corresponds to $f$, is represnted by the following product of $m$ commutators:
$$(*) \quad [a_1,b_1] [a_2,b_2] ... [a_m,b_m]
$$
To visualize that last point is, again, a tool in the classification of surfaces. Take a polygon with $4m+1$ sides and label the sides as follows: leave one side blank; then label the others in order as
$$a_1, b_1, a_1^{-1}, b_1^{-1},...,a_m, b_m, a_m^{-1},b_m^{-1}
$$
Put arrows on the labelled sides to indicate direction: a clockwise arrow for a $-1$ exponent; a counterclockwise arrow otherwise. Now identify sides with matching letters, making sure to match up the arrows. The resulting quotient space is the sphere with $m$ handles and $1$ hole. The labelled sides give the free basis elements of the fundamental group of the quotient surface. The unlabelled side gives you the unique boundary component $f$ of the labelled surface. The polygon itself gives you the path homotopy between $f$ and the commutator $(*)$.
Best Answer
I do not understand a lot of the last paragraph either, but I will give a slightly different proof of the statement based on Bredon, Topology and Geometry, p. 174.
For every point $x\in X$, fix a path $\lambda_x$ from $x_0$ to $x$. Let $\sigma$ be a singular 1-simplex in $X$, i.e. a continuous path $\sigma\colon[0,1]\to X$. Then $\Psi(\sigma):=\overline{\lambda_{\sigma(1)}}\cdot\sigma\cdot\lambda_{\sigma(0)}$ is a loop based at $x_0$, where the dot is juxtaposition of paths and $\overline\lambda(t)=\lambda(1-t)$ is the inverse path. I claim that this induces a well-defined map $\Psi_*\colon H_1(X)\to\pi_1(X,x_0)^{\text{ab}}$, such that $\Psi_*h\colon\pi_1(X,x_0)\to\pi_1(X,x_0)^{\text{ab}}$ is the canonical projection. From this it follows that $h$ induces a monomorphism $\pi_1(X,x_0)^{\text{ab}}\to H_1(X)$.
First well-definedness: Since $\pi_1(X,x_0)^{\text{ab}}$ is abelian, $\Psi$ extends to a homomorphism $\Psi\colon\Delta_1(X)\to\pi_1(X,x_0)^{\text{ab}}$ from the group of singular 1-chains. It is now straightforward to check that boundaries of singular 2-simplices go to zero in $\pi_1(X,x_0)^{\text{ab}}$ (I can elaborate on this if you like), so $\Psi$ induces a homomorphism $\Psi_*\colon H_1(X)\to\pi_1(X,x_0)^{\text{ab}}$ as claimed.
Now if $\sigma$ is a loop based on $x_0$ then $\Psi(h(\sigma))=\overline{\lambda_{x_0}}\cdot\sigma\cdot\lambda_{x_0}$, and the class of this in $\pi_1(X,x_0)^{\text{ab}}$ equals the class of $\sigma\cdot\overline{\lambda_{x_0}}\cdot\lambda_{x_0}\simeq\sigma$, so that $\Psi_*h$ is the canonical projection.