By definition $$[(\alpha *\beta) *\gamma ] (s) = \begin{cases}\alpha (4s) & 0 \leq s\leq \frac{1}{4} \\ \beta(4s-1) & \frac{1}{4}\leq s\leq \frac{1}{2}\\ \gamma(2s-1) & \frac{1}{2}\leq s\leq 1\end{cases}$$
$$[\alpha *(\beta *\gamma) ] (s) = \begin{cases}\alpha (2s) & 0 \leq s\leq \frac{1}{2} \\ \beta(4s-2) & \frac{1}{2}\leq s\leq \frac{3}{4}\\ \gamma(4s-3) & \frac{3}{4}\leq s\leq 1\end{cases}$$
I don't know how to compute the homotopy
algebraically using the following diagram. Specifically how can I find the expressions inside $\alpha ,\beta $ and $\gamma.$(Those circled in red)
Edit: Thanks to Darth Geek my definition was wrong!
The bits in brackets are just a scaling factor, effectively saying "you want to traverse along this loop at a rate of $x$ times faster than the unit rate, and displaced by the relevant amount so that the path starts not at time $t=0$ now, but at the same time that the previous path finished at". In this case, the rate $x$ will be the proportion of the unit length from the left end of the horizontal line at some particular fixed value of $s$ to the right end. Obviously this length is dependent on the value of $s$.
Hopefully that makes sense. So for instance, in a simpler case, if it was just two paths $\alpha$ and $\beta$ glued together along a verticle line in the square diagram at $t=0.5$ then the homotopy would be the trivial
$$H(s,t)=\begin{cases}\alpha(2t) & 0\leq t\leq 1/2 \\ \beta(2t-1)& 1/2\leq t\leq 1\end{cases}$$
Here, we want $H(s,1/2)$ to evaluate to $\beta(0)$, so when $t=1/2$ we want $\beta(f(t))=\beta(0)$ for a function $f$ which we want to determine. We want $f$ to be linear, so we set $f(t)=a+bt$. We want $a+b(1/2)$ to be $0$ and $a+b(1)$ to be $1$, so solving these simultaneous equations we find $f(t)=2t-1$.
Obviously this is an example which doesn't depend on $s$ so it is a bit simpler, but the general method of finding the linear function $f$ is the same - you just have to use some linear geometry to solve a slightly more difficult pair of simultaneous equations.