Your equation $\frac{dy}{dt} = C(t)y$ has the exact solution
$$
y(t) = \mathcal T exp\left(\int_0^t C(s) ds\right) y(0)
$$
where $\mathcal T$ is the time-ordering operator which orders operators at later time to the left. Approximate the integral by a discrete sum ($t=N\Delta t$),
$$y(t) = \mathcal T exp\left(\sum_{i=1}^N H(t_i) \Delta t\right) y(0)$$
and split the terms (*),
$$y(t) = \mathcal T \prod_{i=1}^N exp\left(H(t_i) \Delta t\right) y(0).$$
Now you can trivially apply the time-ordering operator, leading to
$$y(t) = \prod_{i=1}^N exp\left(H(t_i) \Delta t\right) y(0).$$
The latter equation says that one can repeatedly apply matrix exponentials to obtain the solution. This is a fact on which almost any numerical integrator relies.
Now note the step marked by (*). Here, the exponential of a sum of several operators is decomposed into a product of these exponentials, that is, the elementary splitting formula $exp(H_1+H_2) = exp(H_1) exp(H_2)$ is applied several times (the formula is, of course, only an approximation when the two operators do not commute).
How does this help you in solving your equation? In no way. But it's good to know. For the practical part of your question, please read the first answer in your linked question on quora, it basically says it all.