One way to compute the factor group would be to view $\textrm{Im}(\partial_2)$ as the largest element in the following ascending chain:
- $\langle a_0\rangle \subset \langle a_0, 2a_1 - a_0\rangle \subset \langle a_0, 2a_1 - a_0, 2a_2-a_1\rangle \subset \dots \subset \langle a_0, 2a_1 - a_0, \ldots, 2a_n - a_{n-1}\rangle $
Now you can inductively determine the isomorphism class (in the standard classification of finitely-generated abelian groups) of the quotient $\mathbb{Z}^{n+1} / \langle a_0, 2a_1 - a_0, \ldots, 2a_{i+1} - a_{i}\rangle$ for $i = 0, 1, \ldots, n-1$. It is clear, for instance, that $\mathbb{Z}^{n+1}/\langle a_0\rangle \cong \mathbb{Z}^n$. Hopefully that is enough get started.
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.
Best Answer
In this situation the simplices you have from the geometry generate the groups in the chain complex and $\partial_i$ are linear maps between. Therefore, choosing an ordering for each of these generating sets allows you to write $\partial_i$ as matrices, so the problem is reduced to standard linear algebra.
For example, let's call $A,B,C$ the vertices of the original triangle, that are sent to the same point, and $P$ the point in the middle. Then $C_0(M)$ is generate by $\{A,P\}$ (taken in this order), $C_1(M)$ is generate by $\{[A,B],[A,P],[B,P],[C,P]\}$, and finally $C_2(M)$ is generate by $\{[A,B,P],[B,C,P],[C,A,P]\}$.
Your computations, except $H_1$ are indeed correct, but you should justify that $\ker \partial_2=0$.
$\ker\partial_1$ is wrong and you can tell it already because you know that $\operatorname{Im}\partial_2\subset \ker\partial_1$. Indeed, using the basis choosen above we get $\partial_2=\begin{pmatrix} 1 & 1& 1\\ -1 & 0& 1\\ 1 & -1& 0\\ 0 & 1& -1 \end{pmatrix}$ and $ \partial_1=\begin{pmatrix} 0 & -1 & -1 & -1\\ 0 & 1 & 1 & 1 \end{pmatrix}$.
Therefore $\ker \partial_1=\left<\begin{pmatrix} 1 \\ 0 \\ 0\\ 0\\ \end{pmatrix},\begin{pmatrix} 0 \\ 1 \\ -1\\ 0\\ \end{pmatrix},\begin{pmatrix} 0 \\ 1 \\ 0\\ -1\\ \end{pmatrix}\right>$.
Now you need to actually compute the quotient, for example using Smith Normal Form.