That's a pretty big simplicial complex! The standard thing to do - write matrices and try to find the kernel and image - is going to take way too long to be a good use of your time (or mine!) This is one good reason to prefer the language of cellular homology (which requires one knows how to calculate degree effectively), or the language of $\Delta$-complexes, which have uniformly fewer simplices - IIRC the $\Delta$-complex decomposition of the torus requires only two 2-simplices.
Here is a method to progressively simplify this calculation when doing it in practice (other than telling you to find a better method!) It's the matrix discussion in disguise, but I think it's much easier to look at a picture instead of row-reducing.
1) Observe that each diagonal gives rise to two relations: one says that the diagonal is homologous to the horizontal and vertical lines bounding its upper-left, and another says the same for the lower-right. (I am going to ignore signs.) This means that, up to adding boundaries, we may write any term in $C_1$ as a sum of horizontal and vertical lines only.
2) By getting rid of the diagonals, we used up "half" of our 18 relations. What remains now are the following 8 relations, given by adding the relations for each diagonal: if $e_1, \cdots, e_4$ are the edges around one of the squares above, oriented clockwise, then $e_1 + e_2 + e_3 + e_4$ is a boundary. This tells us that if you have one of the edges of a square, up to a boundary you may replace it by minus the sum of the other three.
Use this now to replace any chain by one which has no terms on the central square; this uses up an additional 4 relations (those corresponding to the squares that are on neither of the diagonals, aka the 4 adjacent to the central square). Finally, use this relation on the corner-squares to demand that every chain (up to adding a boundary, as usual) lies on either the boundary, or one of the four vertical "prongs" sticking out. The only remaining relation relates to the center circle,
What we have identified, bit-by-bit, is $C_1(T^2)/\partial C_2(T^2)$, by quotienting the former free abelian group by each relation. We see that it's freely generated by $10$ edges (which makes sense if you happen to know ahead of time the calculation that $H_2(T^2) = \Bbb Z$; if not, we have just proven it by seeing that $17$ of those relations survive and one dies!) Now the map $\partial: C_1(T^2)/\partial C_2(T^2) \to C_0(T^2)$ is transparent:
Because the only edges adjacent to the 4 interior vertices are the 4 prongs, if $[x] \in C_1(T^2)/\partial C_2(T^2)$ has $\partial [x] = 0$, then necessarily the weights on those 4 prongs are zero. Furthermore, you see then that the weights of successive horizontal (or vertical) edges in $[x]$ must be equal. Altogether, you find that $$H_1(T^2) = \text{ker}(\partial) \subset C_1(T^2)/\partial C_2(T^2) \cong \Bbb Z^2,$$ generated by the full horizontal loop and the full vertical loop.
$H_0$ is even easier, so I leave that to you.
Actually, since $\mathbb{Z}$ is a projective abelian group, any exact sequence of the form
$$
0 \to A \to B \to \mathbb{Z} \to 0
$$
splits, although not canonically. If you haven't seen this argument before, just note that for any $b \in B$ mapping to $1$, we can define a map $\mathbb{Z} \to B$ splitting the sequence by sending $n$ to $nb$.
Best Answer
Given a short exact sequence of chain complexes $$0 \to \mathbf A \stackrel{\mathbf i}{\longrightarrow} \mathbf B \stackrel{\mathbf j}{\longrightarrow} \mathbf C \to 0$$ we get a long exact sequence of homology groups $$\ldots \to H_n(\mathbf A) \stackrel{\mathbf i_*}{\longrightarrow} H_n(\mathbf B) \stackrel{\mathbf j_*}{\longrightarrow} H_n(\mathbf C) \stackrel{\partial}{\longrightarrow} H_{n-1}(\mathbf A) \stackrel{\mathbf i_*}{\longrightarrow} H_{n-1}(\mathbf B) \to \ldots$$ See Hatcher Theorem 2.16.
Now recall your question Chain homotopy definition . In the answer we have seen that there is a more general concept of chain complex allowing also non-zero groups in negative dimensions. Theorem 2.16 remains true for short exact sequences of such more general chain complexes; you do not even need to adapt the proof.
An example is the augmented singluar chain complex which has an additional $C_{-1}(X) = \mathbb Z$ (and $C_n(X) = 0$ for $n < -1$). Using this chain complex we get the long exact sequence for reduced homology groups.
The usual Mayer-Vietoris sequence is obtained by applying this to the short exact sequence $$ 0 \to C_*(A \cap B) \stackrel{\varphi}{\longrightarrow} C_*(A) \oplus C_*(B) \stackrel{\psi}{\longrightarrow} C_*^{\mathcal U}(X) \to 0$$ with $\mathcal U = \{A,B\}$ and using the fact that $\iota : C_*^{\mathcal U}(X) \hookrightarrow C_*(X)$ is a chain homotopy equivalence (Proposition 2.21).
To get the Mayer-Vietoris sequence for reduced homology groups we shall show that Proposition 2.21 is also true for the augmented chain complexes (which we denote by $C_\#$); this means that the "augmented map" $\iota : C_\#^{\mathcal U}(X) \hookrightarrow C_\#(X)$ is a chain homotopy equivalence. What does this map look like in dimension $-1$?
Note that $C_0^{\mathcal U}(X) = C_0(X)$ because each singular $0$-simplex automatically maps into some $U_i \in \mathcal U$. Thus $\iota = id$ in dimension $0$ and we can augment $\iota$ with $id: \mathbb Z \to \mathbb Z$ in dimension $-1$.
On p. 123 Hatcher defines a chain map $\rho : C_*(X) \to C_*^{\mathcal U}(X)$ and a chain homotopy $D$ from $\mathbb I$ to $\iota \rho$ and moreover shows that $\rho \iota = \mathbb I$.
We conclude that $\rho = id$ in dimension $0$ because $\iota = id$ in dimension $0$. Therefore we can augment $\rho$ with $id: \mathbb Z \to \mathbb Z$ in dimension $-1$. These augented maps trivially satisfy $\rho \iota = \mathbb I$ on $C_\#^{\mathcal U}(X)$.
We next show that Hatcher's chain homotopy $D$ augments to a chain homotopy from $\mathbb I$ to $\iota \rho$ on $C_\#(X)$; this will prove our modification of Proposition 2.21. Recall that $D_{-1} : 0 \to C_0(X)$ for Hatcher's original $D$; thus $$\partial D_0 = \partial D_0 + D_{-1}\partial = id_0 - \iota_0 \rho_0 .$$ But $\iota_0 \rho_0 = id_0$, thus $\partial D_0 = 0$. Therefore we can augment $D$ as follows to $C_\#(X)$:
Then $\partial D_0 + \bar D_{-1}\varepsilon = 0 = id_0 - \iota_0 \rho_0$ and $\varepsilon \bar D_{-1} + \bar D_{-2}0 = 0 = id_{\mathbb Z} - \iota_{-1} \rho_{-1}$.