Let $Q$ be the quiver $\bullet \to \bullet \to \bullet$. Then $$ \mathbb{C} \xrightarrow{f} \mathbb{C} \xrightarrow{g} \mathbb{C} \quad (1) \\ 0 \xrightarrow{0} \mathbb{C} \xrightarrow{0} 0 \quad (2) $$ are two representations of $Q$, where $f$ and $g$ are two non-zero maps. Is (2) a subrepresentation of (1)?
I am reading the paper. On page 23, the second line, the term $v_{2,-3}$ is not in the bracket of the right hand side of the equation. So it seems that (2) is not a subrepresentation of (1)? The authors use the formula (14) in the end of page 18. Thank you very much.
The question amounts to understanding what a morphism of quiver representations is. If you identify $Q$ with the free category on this quiver, then a representation is just a functor $Q\to \operatorname{mod}\mathbb{C}$. A morphism of representations is then just a natural transformation between these two functors.
Spelling out what this means, a representation of the quiver $\bullet\to \bullet\to\bullet$ is a collection of three vector spaces $U$, $V$ and $W$ together with linear maps $f\colon U\to V$ and $g\colon V\to W$. If you have two such representations $U\stackrel{f}{\to} V\stackrel{g}{\to} W$ and $U'\stackrel{f'}{\to}V'\stackrel{g'}{\to}W'$, then a morphism from the first to the second is three linear maps $\alpha\colon U\to U'$, $\beta\colon V\to V'$, $\gamma\colon W\to W'$ such that the big commutative diagram you can draw commutes, i.e. you have $\beta f=f'\alpha$ and $\gamma g=g'\beta$. The first representation now is (isomorphic to) a subrepresentation of the other if and only if $\alpha, \beta, \gamma$ are injective.
In your case, there exists even no homomorphism from (2) to (1) since in your case $g'=0$ while $g\neq 0$. Hence, if you want $g\beta=\gamma g'=0$, then you need to have $\beta=0$. As $\alpha=\gamma=0$ since they are starting in the $0$-vectorspace it is clear that $\operatorname{Hom}_Q((2),(1))=0$. In particular, (2) cannot be a subrepresentation of (1).
As Hanno pointed out in the comments, (2) is a subrepresentation of a representation $U\stackrel{f}{\to} V\stackrel{g}{\to} W$ iff there is a homomorphism from (2) to this representation (this is clear since (2) is a simple representation, hence any non-zero morphism starting in (2) is a monomorphism) iff $g$ has a non-zero kernel (which you can check easily by writing down what the definition of a morphism of quiver representations is).