Consider the pair $(\Bbb T, A)$ where $A \subset \Bbb T$ is one of the longitudal circles in $\Bbb T$. We thus have the long exact sequence
$$\cdots \to H_{k+1}(\Bbb T) \to H_{k+1}(\Bbb T, A) \stackrel{\partial}{\to} H_k(A) \to H_k(\Bbb T) \to H_k(\Bbb T, A) \stackrel{\partial}{\to} H_{k-1}(A) \to \cdots$$
$(\Bbb T, A)$ has homotopy extension property, as it's a CW-pair. Thus, we have $H_\bullet(\Bbb T, A) \cong H_\bullet(\Bbb T/A) \cong H_\bullet(S^2 \vee S^1)$ which we know is $\Bbb Z$ if $k = 1, 2$ and is trivial otherwise.
The whole point of choosing this pair is that there is a retract $r : \Bbb T \to A$ which, after applying the $H_\bullet$ functor, gives a left-inverse for the map $H_\bullet(A) \to H_\bullet(\Bbb T)$ induced by the inclusion, which proves injectivity of this map. Thus, the snake maps $\partial$ are all zero
Thus, at each $k$, the long exact sequence reduces to the split (the section being $H_\bullet(r)$) exact sequence
$$0 \to H_k(A) \to H_k(\Bbb T) \to H_k(\Bbb T, A) \to 0$$
Thus, $H_k(\Bbb T) \cong H_k(A) \oplus H_k(\Bbb T, A)$ which is $\Bbb Z \oplus \Bbb Z$ for $k = 1$, $\Bbb Z$ for $k = 2$ and trivial otherwise.
To add, $H_\bullet(S^2 \vee S^1)$ can be computed in a similar vein, without using Mayer-Vietoris sequence, by taking the CW-pair $(S^2 \vee S^1, S^2)$. There is the retract $S^2 \vee S^1 \to S^2$ by collapsing the circle to the wedged point, which gives a left-inverse for $H_\bullet(S^1) \to H_\bullet(S^2 \vee S^1)$ induced from inclusion as well as a section for the short exact sequence.
It sounds as if you're confused about the difference between singular/simplicial homology and cellular homology. If you look at the homology chapter of Hatcher's book again, you'll see that these subjects are covered in some detail.
The relevant details are as follows:
The homology of a space is a deep property of that space that is hidden behind a frosted glass wall. That is to say, it should be considered as an extremely fundamental topoogical invariant, but not one which is readily accessible to us. This means that it is difficult to talk about the meaning of the homology groups - there is a nice interpretation in terms of the $n$-dimensional 'holes' in the surface, but that breaks down once we start encountering spaces whose homology groups exhibit non-trivial torsion. This is why we often have to do a lot of messy work in order to get anywhere with homology. Simplicial homology is not a nice piece of mathematics, and it's not obvious why we care about it at all and why we are prepared to invest so much effort studying it. The reason is that it gives us a way to reach through the glass and grasp the homology of the space.
The nice thing is, once we have got a foothold on the homology groups of a few spaces, we can apply various rules again and again to compute many more homology groups, and the whole subject opens up to us.
Pretty soon, we can identify exactly which tools we are using again and again. This leads into the idea of defining a 'homology theory' axiomatically: we just specify that it is a rule that takes a topological space and spits out a sequence of groups in such a way that all our useful tools work. These tools - homotopy invariance, excision, the dimension rule, additivity and the LES of a pair - are known as the Eilenberg-Steenrod axioms.
We then see a reason for computing simplicial homology in the first place - it gives us a model of the Eilenberg-Steenrod axioms. Simplicial homology satisfies the Eilenberg-Steenrod axioms, and so we are guaranteed that A) It is a homotopy invariant and B) we can compute it in a nice way using the other tools. Homotopy invariants that are easy to compute are extremely useful to algebraic topologists!
Where does cellular homology come into this? Well, the other cool thing about homology theories is that they tend to agree. Remember what I said about homology being a deep but somehow 'hidden' feature of a space? Well, simplicial homology is one way to reach into the heart of a space and extract this important invariant. But there are a number of other ways, and it turns out that cellular homology (computing homology from the chain complex of $n$-cells) is one such way. Moreover, cellular homology is much easier to compute than simplicial homology.
The problem is, in order to prove anything about cellular homology, we need to have the theory of homology set up already. So we need some homology theory already set up that satisfies the Eilenberg-Steenrod axioms. That's where simplicial homology (or, more commonly, its cousin singular homology) comes into play. We need to define it first, but once we've used that to set up the theory, we can use it to prove that cellular homology A) satisfies the Eilenberg-Steenrod axioms and B) agrees with our previous definition of homology, and then we can use the (easier) cellular homology to compute homology groups for cell complexes from then on.
Important point (suggested by Mike Miller in the comments): The definition of cellular homology involves picking a choice of cell decomposition for our space. A cell complex may admit many different cell decompositions. How do we know that the cellular homology we get is independent of our choice of decomposition? Why, because we know it is isomorphic to singular homology, which is a topological(/homotopy/weak homotopy) invariant.
Best Answer
$K-x$ 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 $n \ge 3$, $$H_3(X,X-x;\mathbb Z) \approx H_2(\theta;\mathbb Z) \approx 0 $$