When we reduce the order of an o.d.e. by making a substitution of the form $y = \dot{x}$, how do we decide to what space $y$ belongs to? Case in point: consider
$$\ddot{x} = \sin x, \quad x \in \mathbb{S}$$
Introducing $\dot{x}=y$, the above becomes the system
$$\dot{y} = \sin x, \quad \dot{x}=y$$
Question: is $y \in \mathbb{S}$ or $y \in \mathbb{R}$?
Since the above 2nd order ode may physically represent the pendulum, we can say the orbits are on a cylinder $\mathbb{S} \times \mathbb{R}$ and thus $y \in \mathbb{R}$; however if we study the system without any reference to physics, then how do we decide on what space $y$ is defined?
If $x$ is considered to be an element of a manifold $M$ (for example, the circle $S^1$), then an orbit $x(t)$ is a path on $M$ parametrised by the one-dimensional time variable $t \in \mathbb{R}$. For such a situation, you want to understand what $\dot{x}(t) = \frac{\text{d} x}{\text{d} t}(t)$ means, and in particular, in what space it should be defined.
This turns out to be the tangent space of the manifold $M$ at the point $x$, denoted by $T_x M$. So, for any time $t$, the velocity of $x(t)$ at that time, denoted by $\dot{x}(t)$, is an element of $T_{x(t)}M$. For the circle $S^1$, the tangent space at any point on $S^1$ is (isomorphic to) $\mathbb{R}$. This is due to the definition of the tangent space, see here. Such a tangent space is a vector space, so any scalar multiple of a vector is also in that space. In other words, there is no a priori bound on the velocities. This is only natural, as the magnitude of velocities should be imposed by the dynamical system, and not by the underlying phase space. For example, take the ODE $\ddot{x} = \omega^2 x$, with $x \in S^1$. Written as a system, the associated phase space is $S^1 \times \mathbb{R}$, so this ODE lives on the same phase space as your system. However, the velocities have maximal magnitude $\omega$, which I can choose to be as large as I want, whereas the maximal velocity of orbits of your ODE is $|\dot{x}| = 1$.