Which 2-chain has this 1-chain as boundary? Define $\sigma:[0,1] \to \mathrm{R}^2$ by $t \mapsto (t,0)$. And define $\sigma':[0,1] \to \mathrm{R}^2$ by $t\mapsto (1-t,0)$.
Now I want to figure out which 2-chain in $C_2(\mathrm{R}^2)$ has  $\sigma + \sigma'$ as its boundary.
After trying I found a 2-chain $\omega_1 + \omega_2$ defined as in the following picture(we squash two triangles into $[0,1]$ ).

It seems that $\omega_1 + \omega_2$ could have $\sigma + \sigma'$ as boundary. Am I right? Is there any other 2-chain of simpler form satisfying this condition? Actually I find its awkward that we need such a 2-chain to bound a simple 1-chain, or this is something inevitable in singular homology theory?
 A: To supplement the other answer, let me address your question whether there is a 2-chain of simpler form with the same boundary.
We must ask: What's simpler than a sum of two singular 2-simplices?
Not much....
Perhaps just one singular two simplex $\omega : \Delta^2 \to \mathbb R$? But then
$$\partial \omega = \omega \mid [0,1] - \omega \mid [0,2] + \omega \mid [1,2]
$$
cannot equal $\sigma + \sigma'$, because the sum of coefficients of $\sigma+\sigma'$ is even whereas the sum of coefficients of $\partial \omega$ is odd (even after combining terms).
Then you might ask, well, okay, but what about something of the form $n\, \omega$ where $n$ is an even integer? Then we have
$$\partial (n \omega) = n \, \omega \mid [0,1] - n \, \omega \mid [0,2] + n\, \omega \mid [1,2]
$$
In order for this 1-chain to equal $\sigma + \sigma'$, clearly one of $\sigma$ or $\sigma'$ must equal $\omega \mid [0,1]$ and the other equals $\omega \mid [1,2]$ and so
$$\partial(n \omega) = n \, \sigma + n \, \sigma' - n \, \omega \mid [0,2]
$$
Set that equal to $\sigma + \sigma'$ and you get
$$(n-1) \, \sigma + (n-1) \, \sigma' - n \, \omega \mid [0,2] = 0
$$
which is clearly impossible whether $\omega \mid [0,2]$ is equal to one of $\sigma$ or $\sigma'$ or whether it isn't.
A: Technically these are not singular simplexes what you've written. Recall that
$$\Delta^n=\bigg\{(v_1,\ldots, v_n)\in\mathbb{R}^n\ \bigg|\ v_i\geq 0\text{ and }\sum_{i=1}^n v_i=1\bigg\}$$
And so I assume that
$$\sigma,\sigma':\Delta^1\to \mathbb{R}^2$$
$$\sigma(x,y)=(x,0)$$
$$\sigma'(x,y)=(1-x,0)$$
But actually this doesn't matter. I will work with $\mathbb{R}^m$ and any $\sigma$ and $\sigma'$ simplexes from now on, since concrete formulas are unnecessary. I will only assume that $\sigma(0,1)=\sigma'(1,0)$ and $\sigma(1,0)=\sigma'(0,1)$.
WARNING: I deliberately go with formal, explicit definition and not deal with $\Delta^1\simeq [0,1]$ homeomorphism, because that would involve a choice of homeomorphism. Which is important, because orientation is important. That would make calculations more complicated.
Now we want to find two special simplexes $\omega,\omega':\Delta^2\to\mathbb{R}^m$ such that $\partial(\omega+\omega')=\sigma+\sigma'$.
Given $v_1=(1,0,0)$, $v_2=(0,1,0)$ and $v_3=(0,0,1)$ it is enough to define both $\omega,\omega'$ on the boundary $[v_2,v_3]\cup[v_1,v_3]\cup[v_1,v_2]$. That's because we can always extend such continuous function to full simplex by Tietze for example. Note that this is because we work with $\mathbb{R}^m$ in the codomain, otherwise defining on the boundary only would not be enough. And actually this is what makes this entire theory useful: in a sense it checks whether a map defined on the boundary can be extended to full simplex, i.e. whether there are holes! And $\mathbb{R}^m$ does not have holes.
Anyway, lets do some calculations:
$$\partial(\omega+\omega')=\partial\omega+\partial\omega'=$$
$$=\omega_{[v_2,v_3]}-\omega_{[v_1,v_3]}+\omega_{[v_1,v_2]}+\omega'_{[v_2,v_3]}-\omega'_{[v_1,v_3]}+\omega'_{[v_1,v_2]}$$
This gives us a hint: we have to group those partial functions so that they reduce each other leaving $\sigma+\sigma'$ only (pieces of boundaries of $\omega$, $\omega'$ that map to $\sigma$, $\sigma'$ can be chosen arbitrarly, up to a sign). Note that we cannot do that with single simplex, i.e. $\sigma$ is not a boundary of any $\omega$, unless $\sigma(0,1)=\sigma(1,0)$, which we don't assume. And so we have to group between components of $\omega$ and $\omega'$.
So how to do that? Lets try this:
$$\omega(0,y,z)=\sigma(y,z)$$
$$\omega(x,0,z)=z\cdot\sigma(0,1)$$
$$\omega(x,y,0)=y\cdot\sigma(1,0)$$
$$\text{and}$$
$$\omega'(0,y,z)=\sigma'(y,z)$$
$$\omega'(x,0,z)=z\cdot\sigma'(0,1)$$
$$\omega'(x,y,0)=y\cdot\sigma'(1,0)$$
I leave as an exercise that both definitions are correct and yield two continuous maps.
And so $\omega_{[v_2,v_3]}=\sigma$ while $\omega_{[v_1,v_2]}=\omega'_{[v_1,v_3]}$. Analogously $\omega'_{[v_2,v_3]}=\sigma'$ while $\omega'_{[v_1,v_2]}=\omega_{[v_1,v_3]}$. These nicely reduce leaving us with $\partial(\omega+\omega')=\sigma+\sigma'$ only.
Note how we make $1$-simplex from "$\omega$ restricted to one of its faces": via the standard $\Delta^n\to\Delta^{n+1}$ embedding onto one of the faces.
I suppose these formal constructions I've presented here is actually what you had on your mind with the picture. I just gave you tools to do formal verifications.

Actually I find its awkward that we need such a 2-chain to bound a simple 1-chain, or this is something inevitable in singular homology theory?

Yes, raw chain and boundary calculations are tedious. But we rarely do that. The theory is all about the singular homology groups, which once defined have very nice properties making all those calculations unnecessary.
