How to prove "Homotopy is an Equivalence Relation"

This is how it is supposed to be done, yes. You have to go from $0$ to $1$, and by going through the first mapping at twice the speed then going through the second mapping at twice the speed you get the proper speed for the whole composition of maps. (that way you spend half the total time on each)

Once you have shown that if $f\simeq f'$ and $g\simeq g'$, also $gf\simeq g'f'$, you can write $$f_2f_1g_1g_2=f_2(f_1g_1)g_2\simeq f_2 1_Y g_2\simeq 1_Z$$ More generally, the property stated above can be expressed as "composition of homotopy classes is well defined by the formula $[g][f]=[gf]$". This means that the topological spaces with the classes of maps form a category $h\mathbf{Top}$, and two spaces are isomorphic in this category if they are homotopy equivalent. But "isomorphic to" is an equivalence relation.