Start with the time-dependent Schrödinger equation
$$ \hat H \Psi(r, t) = i \hbar \partial_t \Psi (r, t). $$
Using the ansatz $\Psi(r,t) = \psi(r) f(t)$ yields
$$ f(t) \hat H \psi(r) = i \hbar \psi(r) \partial_t f(t)$$
and, via the standard separation of variables trick,
$$ i\hbar\frac{\dot f(t)}{f(t)} = \text{const} = \frac{\hat H\psi(r)}{\psi(r)}; $$
the two sides are equal, but the LHS depends only on $t$ while the RHS depends only on $r$, so they each have to be constant. Let us call this constant $E$. Then, for the time-dependent part, we get
$$ \dot f(t) = -i \frac{E}{\hbar} f(t) $$
Which is manifestly solved by
$$ f(t) = \exp\left\{-i \frac{E}{\hbar} t\right\}. $$
For the spatial part, we find the time-independent Schrödinger equation
$$ \hat H \psi(r) = E \psi(r), $$
which as you observe can be viewed as an eigenvalue equation for $\hat H$. This motivates the choice of name for $E$: Physically, the eigenvalue of the Hamiltonian is the energy.
The "independent" in "time-independent Schrödinger equation" doesn't mean that the wavefunction $\psi(x,t)$ is independent of time, but that the quantum state it defines doesn't change with time.
Since $\psi(x)$ and $\mathrm{e}^{\mathrm{i}\phi}\psi(x)$ for any $\phi\in\mathbb{R}$ define the same quantum state, this does not imply $\partial_t\psi(x,t) = 0$. Indeed, as the solution shows, the time dependence $\mathrm{e}^{\mathrm{i}Et}$ is precisely the kind of dependence that is allowed.
Best Answer
Firstly, there are a few issues with a time-dependent potential, $V(x,t)$. Namely, if we apply Noether's theorem, the conservation of energy may not apply. Specifically, if under a translation,
$$t\to t +t'$$
the Lagrangian $\mathcal{L}=T-V(x,t)$ changes by no more than a total derivative, then conservation of energy will apply, but this resricts the possible $V(x,t)$, depending on the system.
We often treat each Schrödinger equation case by case, as a certain system may lend itself to a different approach, e.g. the harmonic oscillator is easily solved by employing the formalism of creation and annihilation operators. If we consider a time-dependent potential, the equation is generally given by,
$$i\hbar\frac{\partial \psi}{\partial t}=-\frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial \mathbf{x}^2} + V(\mathbf{x},t)\psi$$
Depending on $V$, the Laplace or Fourier transform may be employed. Another approach, as mentioned by Jonas, is perturbation theory, whereby we approximate the system as a simpler system, and compute higher order approximations to the fully perturbed system.
Example
As an example, consider the case $V(x,t)=\delta(t)$, in which case the Schrödinger equation becomes,
$$i\hbar \frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2} + \delta(t)\psi$$
We can take the Fourier transform with respect to $t$, rather than $x$, to enter angular frequency space:
$$-\hbar\omega \, \Psi(\omega,x)=-\frac{\hbar^2}{2m}\Psi''(\omega,x) + \psi(0,x)$$
which, if the initial conditions are known, is a potentially simple second order differential equation, which one can then apply the inverse Fourier transform to the solution.