Sure. Those orientations are fine.
Using a single vertex is a bit shaky...more on this later.
Your matrix looks OK at first glance. rows 3-8 and columns 1-5 give a submatrix whose determinant is obviously nonzero, so its rank is at least 5, and as you observe, the last column is a linear combination of earlier ones, so the rank is exactly 5.
Your computation of $H_1$ is OK, but it's not really a great thing to look at, is it? I mean, is there a $Z/2Z$ factor in there? It's hard to tell.
It turns out that $H_1$ is actually $\Bbb Z \oplus \Bbb Z \oplus \Bbb Z \oplus \Bbb Z$, so let's see how to get there.
From the last item in the quotient (the generator $i - h - a$) we can say that in our group, $i$ is the same as $h+a$, so let's just get rid of it:
\begin{align}
H_1(X, \mathbb{Z})
&= \langle a, b, c, d, e, f, g, h, i \rangle / \langle e-d-c, d-e+f, c-f+g, h-g-b, i-h-a \rangle \\
&= \langle a, b, c, d, e, f, g, h \rangle / \langle e-d-c, d-e+f, c-f+g, h-g-b \rangle \\
\end{align}
and after that, you can do the same with h, and then $g$, then $f$, then $e$:
\begin{align}
H_1(X, \mathbb{Z})
&= \langle a, b, c, d, e, f, g, h \rangle / \langle e-d-c, d-e+f, c-f+g, h-g-b \rangle \\
&= \langle a, b, c, d, e, f, g \rangle / \langle e-d-c, d-e+f, c-f+g \rangle \\
&= \langle a, b, c, d, e, f \rangle / \langle e-d-c, d-e+f \rangle \\
&= \langle a, b, c, d, e \rangle / \langle e-d-c\rangle \\
&= \langle a, b, c, d \rangle \\
\end{align}
at which point the group is evidently the free abelian group on four generators. You can probably, at this point, see how to do all those operations by messing with integer row operations on matrices, but I figured I'd do it out without that.
Back to item 2: what you've got here is not actually a simplicial complex, because each 1-simplex should have as boundary a pair of 0-simplices, but your 1-simplexes all have $v - v$ as their boundaries, and that's not allowed in the definitions.
On the other hand, it all worked out OK, right? How can that be? Well, you've kind of computed the cellular homology of the 2-hold torus, and there's a great theorem that says that this gives the same result as the simplicial homology. But do to it right, you really should turn your octagon into a 16-gon, then put a concentric octagon inside, and a vertex at the very center, and then confirm that every triangle, for instance, has three distinct vertices. Your matrix will be much larger...but the operations on it will go nice and fast and very soon you'll get rid of most of the rows and have something no more complicated than the one you have above.
Best Answer
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.