Hatcher P151 Example 2.47.
We can decompose the Klein bottle K as the union of two Mobius bands $A$ and $B$ glued together by a homeomorphism between their boundary circles.
Then $A$, $B$ , and $A \cap B$ are homotopy equivalent to circles, so the interesting part of the reduced Mayer–Vietoris sequence for the decomposition $K = A \cup B$ is the segment
$$0 \to H_2(K) \to H_1(A \cup B) \stackrel{\phi}{\to} H_1(A) \oplus H_1(B) \to H_1(K) \to 0$$
The map $\phi$ is $\mathbb{Z} \to \mathbb{Z} \oplus \mathbb{Z}, 1 \mapsto (2,-2)$ since the boundary circle of a Mobius band wraps twice around the core circle. Since $\phi$ is injective we obtain $H_2(K) = 0$. Furthermore, we have $H_1(K) \approx \mathbb{Z} \oplus \mathbb{Z}_2$ since we can choose (1, 0) and (1,−1) as a basis for $\mathbb{Z} \oplus \mathbb{Z}$.
So I am really not able to figure out that how can we get the final conclusion that $H_1(K) \approx \mathbb{Z} \oplus \mathbb{Z}_2$ since we can choose (1, 0) and (1,−1) as a basis for $\mathbb{Z} \oplus \mathbb{Z}$? It seems to me that a link is missing between the conclusion that "$H_1(K) \approx \mathbb{Z} \oplus \mathbb{Z}_2$" and the explanation that "we can choose (1, 0) and (1,−1) as a basis for $\mathbb{Z} \oplus \mathbb{Z}$."
I hope the following answers your problem. The exact sequence $$ 0\rightarrow H_2(K)\rightarrow H_1(A\cap B) \xrightarrow{\phi} H_1(A)\oplus H_1(B) \xrightarrow{\psi} H_1(K)\rightarrow 0$$ tells us that $$H_1(K)\cong \left(H_1(A)\oplus H_1(B)\right)/ \ker\psi \cong \left(H_1(A)\oplus H_1(B)\right)/ \operatorname{Im}\phi \cong \left(\mathbb{Z}\oplus\mathbb{Z}\right)/\langle(2,-2)\rangle $$ This follows immediately from the first isomophism theorem, and the fact that $\ker\psi = \operatorname{Im}\phi$. Now we claim that there is an isomorphism of abelian groups $\left(\mathbb{Z}\oplus\mathbb{Z}\right)/\langle(2,-2)\rangle \cong \mathbb{Z}\oplus \mathbb{Z}/2$. To see this, we choose generators $x,y$ for $\mathbb{Z}\oplus\mathbb{Z}/2$ to make the map easier to write down and we look at the map $$ \varphi:\left(\mathbb{Z}\oplus\mathbb{Z}\right)/\langle(2,-2)\rangle\rightarrow \mathbb{Z}\{x\}\oplus \mathbb{Z}/2\{y\}$$ defined by $\varphi(1,-1) = y$ and $\varphi(1,0) = x$, then extended $\mathbb{Z}$-linearly. To check that this is an isomorphism, we must check it is injective and surjective.
To check surjectivity, any element of $\mathbb{Z}\{x\}\oplus \mathbb{Z}/2\{y\}$ is of the form $ax+by$, then $\varphi(a+b,-b) = a\varphi(1,0)+b\varphi(1,-1) = ax+by$.
To check injectivity, suppose we have $\varphi(a,b)=0$. First write $(a,b) = (a+b,0)-(b,-b)$. Then $0=\varphi(a,b) = \varphi(a+b,0)-\varphi(b,-b) =(a+b)x - by$. Hence we must conclude that $-b=2c$ for some integer $c$, and $a=-b$. That is to say, $(a,b) = (2c,-2c)$ for some integer $c$, but this says exactly that $(a,b)\in \langle(2,-2)\rangle$, hence $(a,b)=0$ in the quotient group $\left(\mathbb{Z}\oplus\mathbb{Z}\right)/\langle(2,-2)\rangle$.
The isomorphism $H_1(K)\cong \mathbb{Z}\oplus\mathbb{Z}/2$ may seem a little "artificial", but it is important to remember that the groups $H_1(K)$ are groups generated by actual topological objects: namely, chains from the chain complex. In some sense, the "point" of algebraic topology is to identify these topological objects in terms of abstract groups.