You have a curve $q\colon I \to \mathbb S^3$, where $I$ is a time interval and $\mathbb S^3\subseteq\mathbb R^4$ is the set of all unit quaternions. Here, we represent quaternions as 4-dimensional vectors. Note that if a quaternion should encode an orientation or rotation, it has to be a unit quaternion.
Since $q(t)\in\mathbb S^3\subseteq \mathbb R^4$ lives in the linear space $\mathbb R^4$ you can calculate its time derivative $\dot q(t) = \tfrac d{dt}q(t)$ by
\begin{align}\tag{1}
\dot q(t) = \tfrac d{dt}q(t) = \lim_{h\to0} \frac{q(t+h) - q(t)}h,
\end{align}
meaning that $\bigl(q(t+\delta)-q(t)\bigr)/\delta$ is an approximation to $\dot q(t)$ for small $\delta$. We can see that this approximation lives somewhere in $\mathbb R^4$. More specifically, it is an element of $T_{q(t)}\mathbb S^3$, the tangent space of the sphere at the element $q(t)\in\mathbb S^3$, because $q(\tau)\in\mathbb S^3$ for all $\tau\in I$.
Since $\mathbb S^3$ is a Lie group it is favourable to represent the velocity of $q(t)$ by a vector $\Omega(t)\in\mathbb R^3$ that fulfills
$$\dot q(t) = \frac12 q(t) * \begin{bmatrix}0\\\Omega(t)\end{bmatrix}.$$
This can be thought of as mapping the tangent space $T_{q(t)}\mathbb S^3$ to the tangent space $T_e\mathbb S^3 = \{[0,x^T]^T\in\mathbb R^4\}$ with the neutral element $e=[1,0,0,0]^T$. The $\Omega(t)$ is actually the angular velocity in the body frame. We can calculate $\Omega(t)$ from $\dot q(t)$ by
$$\tag{2}
\Omega(t) = \operatorname{Im}\bigl(2 \overline{q(t)}*\dot q(t)\bigr),
$$
where the overline represents quaternionic conjugation (which is actually the inversion on $\mathbb S^3$) and $\operatorname{Im}$ extracts the imaginary part, hence just drops the first component (which has to be zero here). Now we can put the limit expression for $\dot q(t)$ in here an get
\begin{align*}
\Omega(t) &= \operatorname{Im}\bigl(2 \overline{q(t)}*\dot q(t)\bigr)\\
&= \operatorname{Im}\left(2 \overline{q(t)}*\lim_{h\to0} \frac{q(t+h) - q(t)}h\right) \\
&= \operatorname{Im}\left(2 \lim_{h\to0} \frac{\overline{q(t)}*q(t+h) - \overline{q(t)}*q(t)}h\right) \\
&= \operatorname{Im}\left(2 \lim_{h\to0} \frac{\overline{q(t)}*q(t+h) - e}h\right) \\
&= \lim_{h\to0} 2\operatorname{Im}\frac{\overline{q(t)}*q(t+h)}h,
\end{align*}
since $\operatorname{Im}e = [0,0,0]^T$.
This means that $2\operatorname{Im}\overline{q(t)}*q(t+\delta)/\delta$ is an approximation to $\Omega(t)$, the body frame angular velocity, for small $\delta$.
With equations (1) and (2) from above, you can transform the derivative $\dot q(t) = \tfrac d{dt}q(t)$ into the body frame angular velocity $\Omega(t)$ and back.
Note that $2\operatorname{Im}q(t+\delta)*\overline{q(t)}/\delta \approx \omega(t)$ gives an approximation to the angular velocity with respect to the inertial frame. It simply holds $\omega(t)=\Omega(t)^{q(t)}$, where $x^{q(t)}\in\mathbb R^3$ is the application of the rotation $q(t)$ to $x$ and is defined by
$$ \begin{bmatrix}0\\ x^{q(t)}\end{bmatrix} = q(t)*\begin{bmatrix}0\\ x\end{bmatrix}*\overline{q(t)}.
$$
The simplest way to numerically integrate the attitude is to use a forward Lie-group Euler method. If you somehow calculate $\dot q(t)$ or $\Omega(t)$ you can get the new attitude $q(t+\delta)$ by
$$
q(t+\delta) = q(t)*\widetilde{\exp}(\delta\cdot \Omega(t)),
$$
where $\widetilde{\exp}\colon\mathbb R^3\to\mathbb S^3$ is essentially a Lie group exponential function which is defined by a power series, but can, for unit quaternions, be written in closed form as
$$
\widetilde{\exp}(v) = \cos(\tfrac12 \|v\|) + \frac{v}{\|v\|}\sin(\tfrac12 \|v\|)
$$
for $v\in\mathbb R^3$, where $\|\bullet\|$ is the standard Euclidean norm.
If you're interested in more sophisticated Lie group time integration, you can check out one of my papers [Arnold, Hante 2016] or a nice paper (without quaternions though) of my supervisor [Arnold, Cardona, Brüls 2016] or the preliminary version.
Hope I could clear things up a little.
