What would this homotopy look like? I found a very weird definition of null-homotopy.

Definition.
Let $M$ be a $C^k$ surface which is a subset of $\mathbb{R}^3$. Let $\mathscr{R}$ be an open connected subset of $M$
Let $\gamma$ be a closed curve whose image is the boundary of $\mathscr{R}$
Let $\sigma$ be any closed curve of period $L$ which is either $\gamma$ or lies in $\mathscr{R}$. Let $\sigma(0)=x_0$. Then $\sigma$ is null-homotopic iff there exists a continuous function $\Gamma:[0,L]\times [0,1]\rightarrow M$ such that $\Gamma(t,0)=\sigma(t)$ and $\Gamma(0,s)=\Gamma(t,1)=x_0$, and for all $0<s\leq 1$ and $t\in (0,L)$, $\Gamma(t,s)$ lies in $\mathscr{R}$.

This is an extremely wierd definition. Anyway, how does this homotopy look like? It deforms $\sigma$ to $x_0$ continuously with only one side fixed.
That is, 
What happens at the right side?
Is it possible that a closed curve $\sigma$ becomes a non-closed curve at some time while deforming to $x_0$?

What's written in that box looks like an extremely specialized version, with a few errors, of a very general definition in topology. A lot of what's written can be discarded, and a few corrections can be made, and you get this standard and very general definition: 


*

*In any topological space $X$, a simple closed curve $\sigma : [0,L] \to X$ with $x_0=\sigma(0)=\sigma(L)$ is null homotopic iff there exists a continuous function $\Gamma:[0,L]\times [0,1]\rightarrow X$ such that $\Gamma(t,0)=\sigma(t)$ and $\Gamma(0,s)=\Gamma(t,1)=\Gamma(1,s) = x_0$ for all $0 \le s \leq 1$ and $0 \le t \le L$. 

