$\coth(t)$ has a singularity at $t=0$, so the hypotheses of the existence and uniqueness theorems are not satisfied there. In fact if $\lim_{t \to 0} y(t) = y_0$ and $\lim_{t \to 0} y'(t) = y_1$ exist, $y''(t) \sim (n-2) y_1 t^{-1}$ as $t \to 0$. If $y_1 \ne 0$, this is impossible, as $t^{-1}$ is not integrable at $0$. So there are no solutions with such an initial condition.

EDIT: For some initial conditions at $t=-2$, the solution will "blow up" before $t=-1$.
It suffices to prove, e.g., that on any interval $[-2,a)$ where the solution exists we have $\dfrac{dy}{dt} \ge y^2$ for $t \ge -2$ with $y(-2) > 1$, as then
$$a - (-2) \le \int_{y(-2)}^{y(a)} \dfrac{dy}{y^2} < \int_{1}^\infty \dfrac{dy}{y^2} = 1$$

Now note that if $f = \dfrac{dy}{dt} - y^2$ we have
$$ \dfrac{df}{dt} = \dfrac{d^2y}{dt^2} - 2 y \dfrac{dy}{dt} = ((n-2) \coth(t) - 2 y)(y^2 + f) + \dfrac{n-1}{2} \sinh(2y) $$
Given $n$, there is some $Y$ such that for all $f \in [0,1]$, $t \in [-2,-1]$ and $y \ge Y$,
the right side is positive. So if $y(-2) > Y$ and $y'(-2) > y(-2)^2$, we will have $y' > y^2$ for $t \in [-2,a]$.