Consider the parametrization of the 1-sphere $(\cos s, \sin s)$, $s \in [0,2\pi[$. I understand that the 1-sphere is not contractible.
However, why isn't the map $((\cos s, \sin s), t)$ to $(\cos ts, \sin ts)$ a homotopy of a constant map to the identity map?
On
It's not a homotopy because it's not a well-defined function.
We can compute the 'value' of $((1,0), 1/2)$ in two different ways:
$$ ((1,0), 1/2) = ((\cos(0), \sin(0)), 1/2) \mapsto (\cos(0), \sin(0)) = (1,0) $$ $$ ((1,0), 1/2) = ((\cos(2 \pi), \sin(2 \pi )), 1/2) \mapsto (\cos(\pi), \sin(\pi)) = (-1,0) $$
Since we get two different outputs for the same input, the formula you have given does not define a function.
Your map is not continuous. Remember, $s = 0$ is supposed to be identified with $s = 2\pi$. But $(\cos(t0),\sin(t0))$ is not equal to $\cos(t2\pi),\sin(t2\pi))$ for all $t \in [0,1]$.