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.
Unfortunately the notion of a chain complex is not a standardized concept in the literature. Some authors define it as Wikipedia in your above link. Other authors (for example Allen Hatcher in his book "Algebraic Topology") define it as a sequence
$$\ldots C_n \stackrel{\partial_n}{\to} C_{n-1} \to \ldots \to C_1 \stackrel{\partial_1}{\to} C_0 \stackrel{\partial_0}{\to} C_{-1} = 0$$
Let us denote such objects ad hoc as chain complexes terminating at a dimension $k$. The Wikipedia-definition produces chain complexes terminating at dimension $0$, Hatcher's definition produces chain complexes terminating at dimension $-1$ with the special feature that $C_{-1} = 0$.
However, most authors define it as in this Wikipedia-article as a sequence of $C_n$ and boundary operators $\partial_n : C_n \to C_{n-1}$ with $n \in \mathbb Z$. Let us call this ad hoc an unterminating chain complex. Clearly the latter is a more general concept: We can regard each chain complex terminating at dimension $k$ as an unterminating chain complex by setting $C_n = 0$ for $n <k$ and $\partial_n = 0$ for $n \le k$.
You are right, if we work with a chain complex terminating at dimension $0$, then there is no $\partial_0$. This shows that this concept is inadequate for the purposes of homology. But you can take Hatcher's concept. In fact, you may argue that the singular chain complex of $X$ should be understood as ending at $C_0 = \mathbb Z[Sing_0(X)]$, but can be artificially transformed into a Hatcher-like complex. But formally we can also argue that $Sing_n(X) = \emptyset$ for $n < 0$ so that automatically $C_n = \mathbb Z[Sing_n(X)] = 0$ for $n < 0$. This gives you an unterminating chain complex.
I recommend to work with unterminating chain complexes because also Hatcher-like chain complexes have a serious weakness: Usually one introduces the reduced homology groups by considering the augmented chain complex
$$\ldots \mathbb Z[Sing_1(X)] \stackrel{\partial_1}{\to} \mathbb Z[Sing_0(X)] \stackrel{\epsilon}{\to} \mathbb Z \to 0$$
which terminates at dimension $-2$. This is no longer a Hatcher-like chain complex, but again it can be regarded as an unterminating chain complex.
Best Answer
First of all, you have an error in your computation of $H_0$: The image of $\partial_1$ isn't all of $\mathbb Z^4$; it is only the elements where the sum of the coefficients is zero. Note that this is true of the image of every one-simplex, thus of the entire image. In general, $H_0$ contains a copy of $\mathbb Z$ for every path-connected component of the space. Here, thus $H_0(PB)\cong \mathbb Z$
To figure out the boundary of a $2$-cell, you just traverse the boundary clockwise*, and count how many times you cross each edge, plus one if the edge is oriented clockwise and minus one if it is oriented the other ways. So, in this case, you get $f-a+e-d$. (I think you might be confused on definitions here - it looks like maybe you're trying to treat PB as if it were a simplicial complex - which it is not unless you subdivide the square $U$ into two triangles. You need to treat it as a CW-complex if you include a square)
(*It doesn't really matter which way you go, if there are no $3$-cells; if there are cells of higher dimension, you just have to be consistent about how each cell is oriented)
To compute $H_1$, you note that the image of $\partial_2$ is just multiples of $f-a+e-d$. The kernel of $\partial_1$ is a bit trickier to deal with, but note that this is just sums that have equally many incoming and outgoing edges to each vertex; that is, we get four equations: $$b-e-a=0$$ $$e+d-g-c=0$$ $$c+g-d-f=0$$ $$f+a-b=0$$ You can check that, for any choice of $(a,b,c,d)\in\mathbb Z^4$, there is exactly one element of $(a,b,c,d,e,f,g)\in \mathbb Z^7$ which satisfies all of these equations. Additionally, one can find that there actually is a basis for $\ker \partial_1$ containing $f-a+e-d$ (since this element is "primitive", meaning it is not a non-trivial multiple of any other element of $\mathbb Z^7$ - or, equivalently, that the GCD of its coefficients is $1$), you see that $H_1(PB)\cong \mathbb Z^3$, since it is $\mathbb Z^4$ mod one the copy of $\mathbb Z$ generated by one of its basis elements.
Then, note that $H_2$ is trivial, since $\partial_2$ has trivial kernel.
You might also note that your space is homotopy equivalent to a wedge of three circles, since you can collapse the region $U$ onto the edges $e$ and $d$ and $f$ by a deformation retract. This geometric reasoning makes computing the homology groups way easier.
In general, the way you compute the boundary of an $n$-cell $A$ in a CW-complex $X$ is to consider that it's boundary is identified with $S^{n-1}$. For any $(n-1)$-cell $B$, we can consider a quotient map $q:X\rightarrow X/(X\setminus \operatorname{int}(B))$ - that is, the map collapsing everything outside of $Y$ to a point - which is also identified with $S^{n-1}$. Then, you get a map $S^{n-1}\rightarrow S^{n-1}$ by composing the inclusion of the boundary of $A$ into $X$ with the quotient $q$. The degree of this map tells you the coefficient of $B$ in $\partial_n(A)$. In two dimensions, this is basically the definition I gave earlier, but in higher dimensions, it clarifies what is meant by "orientation".