Kets represent a configuration of the field (and the terminology is usually upgraded from "Hilbert space" to "Fock space"). $\lvert 0 1 0 \cdots 0\rangle$ is that there is 1 excitation (a particle) in some place specified by the second place in that vector.
To explain, let's take a simple example: A particle in a box. Normally, in undergrad quantum you solve the Schrödinger equation:
$$ E_1 \psi_1(x) = - \partial_x^2 \psi_1(x), \quad \psi(0)=0=\psi(L) $$
Here $\lvert \psi_1\rangle$ is a well-defined object in single particle quantum mechanics (i.e. $\psi_1(x) = \langle x | \psi_1 \rangle$). If I solve this eigenvalue equation for the first one: $E_1 = \pi/L$ then I get the wave function $\psi_1(x) = \sqrt{\frac2L}\sin(\pi x/L)$. But I want more particles and it is convenient to think of $\psi_1(x)$ as a field now.
So the way I do this is that I think of a larger "Fock" space that also takes into account number of particles.
If we say these are bosons, then we use boson ladder operators to create a particle. So, say that $a_x^\dagger$ creates a boson exactly at position $x$ in our system. Well, the $\psi(x)$ in our infinite square well is not at a single position, it is spread out over the whole box. So we begin with a state $\lvert{0}\rangle$ that has no particles, and to create the state associated with our solution to Schrödinger's equation we write
$$ \lvert{1 0 0 \cdots}\rangle = \int_0^L dx \,\psi_1(x) a_x^\dagger \lvert{0}\rangle $$
The notation $ \lvert{1 0 0 \cdots}\rangle $ is now that I have one particle in the first energy level. In fact, the boson operator that creates a particle at any energy $n\pi/L$ is just
$$ a_n^\dagger = \int_0^L dx \,\psi_n(x) a_x^\dagger $$
where $\psi_n(x)$ has energy $n\pi/L$. You can easily verify that these are still well-defined boson ladder operators and $[a_n,a_m^\dagger]=0$ if $n\neq m$ (the magic of orthogonality!).
So it is in this sense that the kets are now just "field configurations": In the way I wrote the ket above for the second quantized square well $\lvert 1 2 5 0 0 \cdots \rangle$ means 1 particle in $\psi_1$, 2 particles in $\psi_2$ and 5 particles in $\psi_3$. People like to write this ket as
$$\lvert 1 2 5 0 0 \cdots \rangle = a_1^\dagger (a_2^\dagger)^2 (a_3^\dagger)^5 \lvert \vec{0}\rangle,$$
where the $a$'s are my field operator. Those are what ket's in quantum field theory mean.
The indices inside my ket represent a particular basis for my boson field operators $a_n$ (the energy basis). I could also write a similar quantity in position space basis with $a_x^\dagger$ and those represent the field in quantum field theory $\Psi(\mathbf r)$.
But how does the Hamiltonian change? Well, consider this object:
$$ \mathcal H = \int dx \, a_x^\dagger (-\partial_x^2) a_x. $$
Now apply this to our ket $\lvert{10\cdots}\rangle$. After some boson algebra we get
$$ \mathcal H \lvert{10\cdots}\rangle = \int_0^L dx \,(-\partial_x^2\psi_1(x)) a_x^\dagger \lvert{0}\rangle = E_1 \int_0^L dx \,\psi_1(x) a_x^\dagger \lvert{0}\rangle = E_1 \lvert{10\cdots}\rangle. $$
We have found our Hamiltonian for the second quantized space of kets! Now, people like to work with the field instead of with the kets, so people will use this Hamiltonian $\mathcal H$ with the field operator $a_x$ to obtain its full evolution with the Heisenberg picture.
