To answer your specific questions:
Now is this Hamiltonian operator supposed to be applied in the
Schrodinger's equation in QM?
Yes. This operator describes the evolution of the quantum state in exactly the same way you are used to. Namely, the state at any particular time is a vector in a Hilbert space, say $\lvert\text{state}\rangle$, and the state some time $t$ later is $e^{-iHt}\lvert\text{state}\rangle$. This raises the question...
What is the vector space this Hamiltonian operator is going to act
upon?
In general, the Hilbert space of a QFT is the complex span of the space of field configurations. For example, for a real scalar field, the field configurations are all the functions from space $\mathbb{R}^d$ (not spacetime, just space) to $\mathbb{R}$. Symbolically the set of field configurations $B$ is
$$B=\{\phi\,|\,\phi:\mathbb{R}^d\rightarrow\mathbb{R}\}.$$
Now take $B$ to be the formal basis for a vector space $\mathcal{H}$. This is the Hilbert space of QFT. So if $\phi_1$ and $\phi_2$ are two different functions from $\mathbb{R}^d$ to $\mathbb{R}$, the Hilbert space will include states like $\vert\phi_1\rangle$, $\vert\phi_2\rangle$, and $\alpha\vert\phi_1\rangle+\beta\vert\phi_2\rangle$. (Note that it is not the case that $\alpha\vert\phi\rangle=\vert\alpha\phi\rangle$, and it is not the case that $\vert\phi_s\rangle+\vert\phi_2\rangle=\vert\phi_1+\phi_2\rangle$. The linear combinations like $\alpha\vert\phi_1\rangle+\beta\vert\phi_2\rangle$ are formal. Also note that we take the different elements of $B$ to be formally orthogonal. So if $\phi_1\neq\phi_2$ we have $\langle\phi_2\vert\phi_1\rangle=0$.) This Hilbert space $\mathcal{H}$ is indeed the Hilbert space that the Hamiltonian operator acts on. So, for example, at some time the state of the universe might be $\alpha\vert\phi_1\rangle+\beta\vert\phi_2\rangle$. Then the state of the universe some time $t$ later will be
$$
e^{-iHt}(\alpha\vert\phi_1\rangle+\beta\vert\phi_2\rangle).
$$
Just like you're used to.
(One might consider having a Hilbert space of this form as a definition of what a QFT is. It's in the name after all: a quantum field theory is just a quantum theory where the states are superpositions of field configurations, rather than, say, superpositions of particle configurations. All the other objects/properties that commonly get talked about in a QFT course, like Lagrangians, Lorentz symmetry, etc. are all just extras. There are indeed proper QFTs without Lagrangian formulations, or without Lorentz symmetry, and so on.)
When/How are the particle creation-annihilation process going to come
into the picture?
We now have a Hilbert space $\mathcal{H}$, and we have a basis for it, $B$. As with any vector space there are many choices of basis for $\mathcal{H}$. The basis $B$ turns out to not be the only (or even most) useful basis. Remember that in one-particle QM, alongside the position basis $\{\vert x\rangle\}_{x\in\mathbb{R}}$, a common basis for the Hilbert space is the harmonic oscillator eigenstate basis: $\{\vert0\rangle,a^\dagger\vert0\rangle,a^\dagger a^\dagger\vert0\rangle,\ldots\}$. In QFT one often talks about the "Fock space" basis, which is analogous to the harmonic oscillator eigenstate basis you're familiar with from one-particle QM.
The elements of $B$ have the physical interpretation of field configurations. The elements of the Fock basis, on the other hand, have the physical interpretation of particles. These two bases for $\mathcal{H}$ are, of course, related by something like a unitary transformation. So states from the Fock basis like $a^\dagger_p a^\dagger_q\vert 0\rangle$ can be written as a "sum" of field configuration states like $\vert \phi\rangle$. And field configuration states like $\vert \phi_1\rangle$ or $\alpha\vert\phi_1\rangle+\beta\vert\phi_2\rangle$ can be written as "sums" of Fock basis states. In practice, the way to go back and forth between these two bases is via the relation
$$
\hat{\phi}(x)=\int\!\frac{\mathrm{d}^dp}{(2\pi)^d}\,\frac{1}{\sqrt{2\omega_p}}(a_pe^{-ip\cdot x}+a^\dagger_pe^{ip\cdot x}),
$$
where $\hat{\phi}(x)$ are the field operators, the operators of which the elements of $B$ are eigenstates. (e.g. the operator $\hat{\phi}(x)$ acting on $\vert\phi_1\rangle\in B$ gives
$
\hat{\phi}(x)\vert\phi_1\rangle=\phi_1(x)\vert\phi_1\rangle.
$)
Realize that the above is all just a rough sketch. But it's the sketch you should be holding in your head when learning about QFT. Now for some editorializing. Many textbooks and courses do a bad job of explaining these fundamentals. In fact, QFT pedagogy is rife with such bad concepts as "second quantization" and false statements like "QFT is QM made compatible with special relativity," "The Klein-Gordon and Dirac equations are relativistic versions of the Schrodinger equation," "In QFT we use the Heisenberg equation, not the Schrodinger equation," "We replace the wavefunction from QM with the field operator," "There are no wavefunctions in QFT," and a million others.
