You're basically asking how to find the equation to the equation
$$
\frac{d\mathbf{R}}{dt} = M(t) \mathbf{R}.
$$
It can be shown that the answer can always be written as
$$
\mathbf{R}(t) = U(t) \mathbf{R}(0),
$$
for a linear operator $U(t)$, but the form of $U(t)$ depends on whether $M(t_1)$ commutes with $M(t_2)$ for all $t_1$ & $t_2$. If $[M(t_1), M(t_2)] = 0$, then it can be shown that we have
$$
U(t) = \exp \left[ \int_0^t dt'\, M(t') \right]. \tag{1}
$$
In particular, the case where $M$ is independent of $t$ means that $U(t) = e^{M t}$.
If, on the other hand, $[M(t_1), M(t_2)] \neq 0$, then we have to define the evolution operator in terms of a Dyson series:
$$
U(t) = 1 + \sum_{n = 1}^\infty \int_0^t dt_1 \int_0^{t_1} dt_2 \cdots \int_0^{t_{n-1}} dt_{n} \, M(t_1) M(t_2) \cdots M(t_n) \tag{2}
$$
In other words, this is an infinite sum whose first term is a single integral, whose second term is a double integral, whose third term is a triple integral, etc. This turns out to be equivalent to
$$
U(t) = 1 + \sum_{n = 1}^\infty \frac{1}{n!} \int_0^t dt_1 \int_0^t dt_2 \cdots \int_0^t dt_{n} \, \mathcal{T} \left[ M(t_1) M(t_2) \cdots M(t_n) \right]
$$
where $\mathcal{T} \left[ M(t_1) M(t_2) \cdots M(t_n) \right]$ is the time-ordered product of the operators, i.e., the operators are reordered so that the arguments decrease from left to right in the string. In this form, it is a bit easier to see how this might be equivalent to the form (1) above.
In particular, if we are talking about 2-D rotations, then the 2-D rotation group is Abelian, i.e., every rotation commutes with every other rotation and we're in the case where we can just write $U(t)$ in the form from (1) above. If, on the other hand, we're talking about 3-D rotations, then it is no longer the case that two rotations at different times will commute (because 3-D rotations generally do not commute), and so we would have to deal with the form (2) for $U(t)$ instead.
