Yes, you are on the right track. The series you have there is called Dyson's series.
First note that the $n$'th term looks like
$$
U_n = \left(-\frac{i}{\hbar}\right)^n\int_0^t dt_1 \cdots\int_0^{t_{n-1}} dt_{n} H(t_1)\cdots H(t_n)
$$
The order of the Hamiltonians is important, since we work with operators. Each term in the series possess a nice symmetry, allowing us to write:
\begin{align}
U_n = \left(-\frac{i}{\hbar}\right)^n \int_0^t dt_1 \cdots\int_0^{t_{n-1}} dt_{n}\ H(t_1)\cdots H(t_n) =
\frac{\left(-\frac{i}{\hbar}\right)^n}{n!}\int_0^t dt_1 \cdots\int_0^t dt_{n} \mathcal{T}\left[H(t_1)\cdots H(t_n)\right]
\end{align}
Two things happened: first, we "overcount" by making the upper limits equal to $t$ on all the integrals. This is compensated by the factor of $\frac{1}{n!}$. You'll need to convince yourself why this factor is needed ;)
Second, by this change of integration area we mess up the ordering of the Hamiltonians in the process. This is where the time-ordering symbol $\mathcal{T}$ comes in. Basically, this operator ensures that the Hamiltonians are always ordered in the correct way. For instance for $n=2$ it operates as
\begin{align}
\mathcal{T}[H(t_1) H(t_2)] = \begin{cases}
H(t_1) H(t_2) & t_2 > t_1\\
H(t_2) H(t_1) & t_2 < t_1
\end{cases}
\end{align}
Putting everything together we have
$$
U(t, t') = 1 + \sum_{n=1}^\infty \frac{\left(-\frac{i}{\hbar}\right)^n}{n!} \int_{t'}^t dt_1 \cdots\int_{t'}^t dt_n \mathcal{T}[H(t_1)\cdots H(t_n)]
$$
Frequently, this is denoted symbolically as
$$
U(t, t') = \mathcal{T}\exp\left(-\frac{i}{\hbar} \int_{t'}^t H(t_1) dt_1\right)
$$
This notation is understood as representing the power series.
From the given information it's a little hard to deduce what you want. But say you have a time dependent magnetic field interacting with a system of particles. Then the spin state of the system of the particles will be time dependent. Therefore, the Hamiltonian of the system of particles will be time dependent. Usually a time dependent Hamiltonian comes from a non-conserved system (here we observe only the system of particles interacting with the magnetic field, rather than the whole picture, which would also include what is generating the magnetic field). Since we ignore what is generating the magnetic field, then the system is not conserved therefore the Hamiltonian is often time dependent.
That's why it is used without hesistation. It's all about the context of the problem.
Best Answer
The short answer is that the time evolution operator for a time-dependent Hamiltonian has two times, the initial and final $U (t,s) $.
Therefore defining $A (t,s)=U (s,t)A U (t,s) $, the Heisenberg equation is obtained differentiating with respect to $t $. Schroedinger equation is obtained differentiating $U (t,s) \psi (s)$ instead. The two are equivalent in the usual sense, i.e. they both give the same time-evolved transition amplitudes.