Here, $M$ is a smooth manifold and $V(M)$ is the space of vector fields on $M$. Now for the definition of "flow, it is said in the picture above that $\sigma(t_2, \sigma(t_1,x))=\sigma(t_2+t_1,x)$. Is it abbreviated that $\sigma=\sigma_X$ in this relation? Also, is this relation defined to hold for flows? Or can I derive this property from the definition of flows? Could anyone please explain?
Let $M$ be a manifold and $X$ be a vector field on $M$, the flow of $X$ is denoted by $\sigma_X$ or simply $\sigma$, it is defined so that for all $x\in M$, $\sigma(\cdot,x)$ is the maximal solution of: $$\dot{c}=X\circ c,c(0)=x\tag{1}.$$ Intuitively, $\sigma(t,x)$ is obtained from $x$ following the maximal integral curve of $(1)$ during a time $t$.
Then, it can be derived that $\sigma(\cdot,x)$ is a one parameter subgroup of $\operatorname{Diff}(M)$, this is the composition relation flows naturally satisfy, it derives from the geometrical interpretation/definition.
The point $\sigma(t+t',x)$ is obtained from $x$ following the maximal integral curve of $(1)$ during a time $t+t'$. Or, since $(1)$ is autonomous (time-independent) from $x$ following the maximal integral curve of $(1)$ during a time $t$, then following it from $\sigma(t,x)$ during a time $t'$. Said differently, you can travel from $x$ to $\sigma(t+t',x)$ along the maximal integral curve of $(1)$ and choose to make a stop at $\sigma(t,x)$. With symbols: $$\sigma(t+t',x)=\sigma(t,\sigma(t',x)).$$
If you want a formal proof with computations, here it is: $\sigma(\cdot,\sigma(t',x))$ and $\sigma(\cdot+t',x)$ are both solutions of: $$\dot{c}=X\circ c,c(0)=\sigma(t',x),$$ indeed the time-derivative of $\sigma(\cdot+t',x)$ at $t$ is $X(\sigma(t+t',x))$ and $\sigma(0+t',x)=\sigma(t',x)$.