Pretty much everything you've said is correct. Because the boundary of an n-simplex is defined to be a sum of the restrictions of the map to various "faces" of the $n$-simplex, each of which is therefore an $n$-simplex...then if there's only one $0$-simplex, the boundary of the 1-simplex has to be $\sigma_0 \pm \sigma_0$. (And the sign is determined by the definition.)
Regarding $\sigma_2$ and orientation: No, it's not possible to orient it so that it gives a negative sign. The simplex $\sigma_2$ is a particular map from $\Delta^2$ (which I like to think of as $\{(x,y,z) \mid 0 \le x,y,z \le 1, x + y + z = 1 \}$) to the single point (call it $P$) defined by $(x, y, z) \mapsto P$. That's all it is. There's no "orientation" of the map.
It's possible to choose a different generator for $C_2$, namely $-\sigma_2$. But that doesn't change $\sigma_2$ itself.
Otherwise...you've got it exactly right.
Singular homology appears, at first, to be just weird. If you do simplicial homology, everything is nice and finite and manageable, so why go to the set of all possible continuous maps? Well you might ask. One answer is that after you prove a few more things, you'll find that the two techniques yield exactly the same results for simplicial complexes, so it doesn't matter. But that doesn't address "Why make it so complicated???" A good answer to that is "because not everything whose homology you wish to compute is actually as nice as a simplicial complex," so you either have to go through the (complicated, to me at least) simplicial approximation theorem, or you have to just say that this singular homology isn't looking so bad after all. And after you've done a few examples like this one, and the homology of a circle, and a few others, you rapidly learn how to do stuff where you never need to think about the singular simplices themselves, just as when you learn that an ordered pair is defined by
$$(a, b) := \{ \{a\}, \{a, b\}\},
$$
it seems really awkward and messy. But then you prove that $(a, b) = (c, d)$ if and only if $a = c$ and $b = d$, and thereafter you use only that lemma, and never look at the formal set-theory definition again.
It's possible I screwed something up here, but I think this construction works for showing that a path plus its reverse is homologous to a constant, and is simpler than Max's more general answer to showing that in general $\alpha + \beta$ is homologous to $\alpha * \beta$ for any composable paths.
Represent points in $\sigma_2$ by their barycentric coordinates, i.e. they are points $(t_0, t_1, t_2)\in [0,1]^3$ such that $t_0 + t_1 + t_2 = 1$. Then define $\varphi_2\colon \sigma_2 \to X$ by
$$ \varphi_2(t_0, t_1, t_2) = \varphi_1(t_1, 1 - t_1) $$
where $\varphi_1$ is your singular $1$-simplex.
Then by definition of the boundary operator $\partial \varphi_2 = \partial_0\varphi_2 - \partial_1\varphi_2 + \partial_2\varphi_2$ where $\partial_i\varphi_2$ is $\varphi_2$ precomposed with the face map $l^i\colon \sigma_1 \to \sigma_2$, which are explicitly given by $l^0(s_0, s_1) = (0, s_0, s_1)$, $l^1(s_0, s_1) = (s_0, 0, s_1)$, and $l^2(s_0, s_1) = (s_0, s_1, 0)$. Therefore we can compute $\partial_0\varphi_2(s_0, s_1) = \varphi_1(s_0, 1 - s_0)$ which is just $\varphi_1$; $\partial_1\varphi_2(s_0, s_1) = \varphi_1(0, 1)$ which is constant and hence in $B_1$ by your exercise; and $\partial_2\varphi_2(s_0, s_1) = \varphi_1(s_1, 1-s_1) = \varphi_1(1-s_0, s_0)$ which is $\bar{\varphi}_1$. That is,
$$ \partial\varphi_2 = \varphi_1 - c + \bar{\varphi_1} \in B_1 $$
Best Answer
I can give you the intuition behind the answer and someone else can fill in the exact details but here is the gist of it. Say you have some $1$-simplex $\phi$ in your space $X$:
You should picture $\phi_1$ and $\phi_2$ as the two $1$-simplices which join together to form $\phi$ like this:
Then if we can find a $2$-simplex $\sigma: \Delta^2 \rightarrow X$ such that $\partial_1 \sigma = \phi$, $\partial_0 \sigma = \phi_1$, $\partial_2 \sigma = \phi_2$ then it's easy to check that $\partial \sigma = \phi_1 + \phi_2 - \phi$. We construct $\sigma$ as the composition of these functions:
You can intuitively see that $\sigma$ fits our criteria. Restricting $\sigma$ to the appropriate boundaries gives the maps we wanted and I have coloured in the appropriate colours to make this easier to see.
I'm pretty sure that $f$ is given by $(t_0, t_1, t_2) \mapsto (1-t_0 - (1-a)t_1,t_0 + (1-a)t_1)$ but can not verify it for myself so not 100% certain