Lel $I:=[0,1]$ and consider the Moebius strip $M:=I\times I/\sim$ defined quotienting $I\times I$. If I consider the boundary of $M$, we call it $B$, then $B$ is parametrized by the map $\pi\circ \beta$, where $\pi$ is the quotient map and the other map is defined as follows
\begin{align*} \beta\colon I & \longrightarrow I\times I\\ t &\longmapsto \bigg\{ \begin{array}{ll} (1,2t) & \mbox{if } 0\leq t<1/2 \\ (0,2t-1) & \mbox{if } 1/2\leq t\leq 1 \end{array}. \end{align*} Taking $(0,0)$ as base-point, I want prove that there exist a base-point preserving homotopy between the loop $\pi\circ \beta$ and the loop defined by \begin{align*} \mu\colon I & \longrightarrow I\times I/\sim\\ t &\longmapsto \bigg\{ \begin{array}{ll} \pi(2t,2t) & \mbox{if } 0\leq t<1/2 \\ \pi(2t-1,2t-1) & \mbox{if } 1/2\leq t\leq 1 \end{array}. \end{align*} Any help?
You've almost written it yourself! If you let $s$ range from $0$ to $t$ and define
\begin{align*} H(s, t) &\longmapsto \bigg\{ \begin{array}{ll} (2s,2t) & \mbox{if } 0\leq t<1/2 \\ (2s-1,2t-1) & \mbox{if } 1/2\leq t\leq 1 \end{array}. \end{align*}
then $\pi \circ H$ defines a homotopy (if I've got your identifications right). but since we usually like for $s$ to go from $0$ to $1$, we can define
\begin{align*} K(s, t) &\longmapsto \bigg\{ \begin{array}{ll} (2st,2t) & \mbox{if } 0\leq t<1/2 \\ (2st-1,2t-1) & \mbox{if } 1/2\leq t\leq 1 \end{array}. \end{align*}
and this gives (I think!) the desired homotopy.