= Solution
For $t<1/2$, the number $f(t)$ is negative real, so the proposed identity would force
$$
\phi(t)\in(2\mathbb Z+1)\pi.
$$
Continuity on the connected interval $[0,1/2)$ makes $\phi$ equal to one fixed odd multiple of $\pi$ there. For $t>1/2$, the number $f(t)$ is positive real, so $\phi$ must instead equal one fixed even multiple of $\pi$. These two constants cannot agree, and their one-sided limits at $1/2$ therefore contradict continuity. The fact that $f(1/2)=0$ removes the phase condition at that one point but does not repair the discontinuity.
Now let $g:[0,1]\to\mathbb C\setminus\{0\}$ and put $h=g/|g|$. This is a continuous path in the <unit circle>. The exponential map $\theta\mapsto e^{i\theta}$ is a <covering space>[covering map], so the <path lifting theorem> supplies a continuous lift $\theta:[0,1]\to\mathbb R$ after one value of $\theta(0)$ is chosen. Hence
$$
g(t)=|g(t)|e^{i\theta(t)}.
$$
One can obtain the same lift directly from the allowed special case: by uniform continuity, subdivide $[0,1]$ so that on each subinterval $h(t)/h(t_j)$ lies in the right half-plane, choose its continuous local argument there, and add a multiple of $2\pi$ to match the preceding endpoint.
If $\theta$ and $\widetilde\theta$ are two such phases, then
$$
\frac{\theta(t)-\widetilde\theta(t)}{2\pi}\in\mathbb Z.
$$
This integer-valued function is continuous and hence constant. Therefore
$$
r(g)=\theta(1)-\theta(0)
$$
does not depend on the chosen lift.
For $u(t)=g(t^2)$, the phase $\theta(t^2)$ gives
$$
\boxed{r(u)=r(g)}.
$$
For $v(t)=g(t)^2$, the phase $2\theta(t)$ gives
$$
\boxed{r(v)=2r(g)}.
$$
Finally, take
$$
g_1(t)=1,
\qquad
g_2(t)=e^{2\pi it}.
$$
They have the same two endpoints, but $r(g_1)=0$ and $r(g_2)=2\pi$. This endpoint discrepancy records the <winding number> of the closed path.
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