Let me explain this in detail for you.
Firstly, the boundary of a $2$-simplex $\sigma$ is a simplicial complex $K$. An abstract simplicial complex to which the vertex scheme of $K$ is isomorphic is $\{\{v_0\},\{v_1\},\{v_2\},\{v_0,v_1\},\{v_0,v_2\},\{v_1,v_2\}\}$; the geometric realization of this abstract simplicial complex is $K$.
Let us understand the simplicial chain complex $C(K)$. The chain group in dimension $0$, $C_0(K)$, is the free abelian group on the set of vertices of $K$ (by definition), the chain group in dimension $1$, $C_1(K)$, is the free abelian group on the set of (oriented) $1$-simplices of $K$, and the chain groups in other dimensions are trivial (there are no simplices of dimension greater than $1$ or less than $0$ in $K$).
In particular, the chain complex of $K$ looks like $\cdots\to 0\to 0\to C_1(K)\to C_0(K)\to 0\to 0\to\cdots$; we need to compute the kernel and image of the map $C_1(K)\to C_0(K)$, i.e., the set of cycles and boundaries in dimensions $1$ and $0$, respectively. Firstly, your intuition of "cycle" should suggest that an oriented sum of the $1$-simplices of $K$ should be a cycle. Indeed, you can check this by direct computation:
$\partial([v_0,v_1]+[v_1,v_2]+[v_2,v_0])$
$=\partial([v_0,v_1])+\partial([v_1,v_2])+\partial([v_2,v_0])$
$=(v_1-v_0)+(v_2-v_1)+(v_0-v_2)$
$=0$.
I leave the following important step as an exercise:
Exercise 1: Prove that if $n_0[v_0,v_1]+n_1[v_1,v_2]+n_2[v_2,v_0]$ is a cycle, then $n_0=n_1=n_2$. (Hint: directly compute the image of the simplicial $1$-chain under the boundary map $C_1(K)\to C_0(K)$.)
Therefore, the kernel of $C_1(K)\to C_0(K)$ is the cyclic subgroup of $C_1(K)$ generated by (the cycle) $[v_0,v_1]+[v_1,v_2]+[v_2,v_0]$. In particular, the first homology group $H_1(K)=Z_1(K)/B_1(K)=Z_1(K)\cong \mathbb{Z}$; there are no boundaries in dimension $1$!
In the case of $H_0(K)$, it is much easier to compute this group directly than to first compute the groups of cycles and boundaries $Z_0(K)$ and $B_0(K)$, respectively, in dimension $0$. Let me explain. The group $Z_0(K)$ is clearly $C_0(K)$ because $C_{-1}(K)=0$ but $B_0(K)$ is a little more difficult to describe. We can instead observe that the group $H_0(K)$ is generated by the cosets of $v_0,v_1,v_2$ in $Z_0(K)/B_0(K)=C_0(K)/B_0(K)$ because $C_0(K)$ is the free abelian group generated by $v_0,v_1,v_2$. In order to understand $H_0(K)$, we need to understand the relations between $v_0,v_1,v_2$. Let us compute boundaries:
$\partial([v_0,v_1])=v_1-v_0$
$\partial([v_1,v_2])=v_2-v_1$
$\partial([v_2,v_0])=v_0-v_2$.
In other words, $v_0+B_0(K)=v_0+(v_1-v_0)+B_0(K)=v_1+B_0(K)$ and similarly $v_1+B_0(K)=v_2+B_0(K)$; i.e., $v_0,v_1,v_2$ all have the same image in $H_0(K)=C_0(K)/B_0(K)$! Therefore, the group $H_0(K)$ is generated by the image of, say, $v_0$.
Exercise 2: Prove that if $nv_0$ is a boundary, then $n=0$. (Hint: let us define the augmentation map $\epsilon:C_0(K)\to \mathbb{Z}$ by the rule $\epsilon(v_i)=1$ for all $1\leq i\leq 3$. Prove that the composition of $\epsilon$ with the boundary map $\partial:C_1(K)\to C_0(K)$ is zero.)
Finally, we have shown that $H_0(K)$ is the infinite cyclic group generated by $v_0$, that is, $H_0(K)=\mathbb{Z}$!
I leave the case of the simplicial complex $L$ consisting of the proper faces of a $3$-simplex as a (more difficult) exercise.
I hope this helps!