Calculation of Local homology groups.$H_*(K,K-x,\mathbb{Z})$? Let $K$ be a simplicial complex and $x \in K$.
Then, how to calculate the relative homology $H_*(K,K-x,\mathbb{Z})$ by
the simplical complex manner?
I am not sure what sub-simplicial complex of $K$ is defined by the notation $K-x$.

Ref
What exactly are the elements of a local homology group?
Local homology of isolated singularity
Local Homology for Graphs
 A: $K-x$ does not represent a subcomplex, it represents $K$ with the point $x$ removed. So $H_*(K,K-x;\mathbb Z)$ does not make sense in simplicial homology. 
On the other hand, even when you use singular homology, you can express the answer using simplicial homology. Here's a way to do that.
Let $\sigma$ be the unique simplex whose interior contains the point $x$. You can assume that $x$ is the barycenter of $\sigma$ (because there is a self-homeomorphism of $K$ preserving every simplex that takes $x$ to the barycenter of $\sigma$). It follows that $x$ is a vertex of every iterated barycentric subdivision of $K$.
Define the star of $x$, denoted $\text{Star}(x)$ to be the subcomplex of $X$ obtained as the union of all simplices of the 2nd barycentric subdivision $K''$ that contain $x$. 
Define the link of $x$, denoted $\text{Link}(x)$, to be the subcomplex of $\text{Star}(x)$ consisting of the union of all simplices of $\text{Star}(x)$ that do not contain $x$. In the example depicted in your post, $\text{Link}(x)$ is a $\theta$ graph.
The basic fact is that $\text{Link}(x)$ is a deformation retract of $\text{Star}(x)-x$, so the singular homology of $\text{Star}(x)-x$ is isomorphic to the homology (singular or simplicial) of $\text{Link}(x)$.
By excision we have $H_*(K,K-x;\mathbb Z) \approx H_*(\text{Star}(x),\text{Star}(x)-x;\mathbb Z)$.
By the long exact sequence of a pair, we have
$$H_*(\text{Star}(x),\text{Star}(x)-x;\mathbb Z) \approx \widetilde H_{*-1}(\text{Star}(x)-x;\mathbb Z) \approx \widetilde H_{*-1}(\text{Link}(x);\mathbb Z)
$$
where $\widetilde H$ represents reduced homology. In this chain of isomorphisms, the term on the left is relative singular homology, the term in the middle is absolute singular homology. The term on the right is any homology you like, for example you could take it to be simplicial homology, because $\text{Link}(x)$ is a simplicial complex.
In your example
$$H_1(X,X-x;\mathbb Z) \approx \widetilde H_0(\theta;\mathbb Z) \approx 0
$$
and
$$H_2(X,X-x;\mathbb Z) \approx H_1(\theta;\mathbb Z) \approx \mathbb Z \oplus \mathbb Z
$$
and, if $n \ge 3$,
$$H_3(X,X-x;\mathbb Z) \approx H_2(\theta;\mathbb Z) \approx 0
$$
