This is a strange question. To begin with there is only one quantum theory. The most comprehensive one is relativistic quantum field theory and the most comprehensive application of its methods is the Standard Model of Particle Physics.
Of course we do not always need the full machinery of relativistic QFT, and most physics problems can only be tackled in approximations adequate to the problem at hand. E.g., much of atomic physics (atoms with not too large ##Z##), molecular physics, and condensed-matter physics can be treated in the non-relativistic approximation.
In non-relativistic physics (e.g., atomic physics) you often have a situation where you deal with a fixed number of particles (like the electrons around the nucleus), i.e., no particle creation and annihilation process (the normal thing in the relativistic realm). Then you can use quantum mechanics.
The name "second quantization" is a bit misleading. It comes from the fact that it is a simple heuristic to consider the Schrödinger equation for one particle ("first quantization") as a classical field theory, formulate it in terms of the Hamilton formalism and than "canonically quantize" it. Then you end up with creation and annihilation operators for fermions or bosons (depending on whether you use anticommutation or commutation relations for the field operators). All this is of course simply a heuristic way to motivate the formalism, and it's didactically not the best way, because it leads to this misunderstanding as if non-relativistic QFT were something else than non-relativistic many-body QM.
Take as the most simple example a spin-0 particle. On basis you can use are the position eigenstates ##|\vec{x} \rangle##. Now spin-0 particles are bosons. Thus a basis for for ##N## indistinguishable bosons is given by the totally symmetrized product states
$$|\vec{x}_1,\ldots,\vec{x}_N \rangle^{+} = \frac{1}{\sqrt{N!}} \sum_{P \in S_N} |\vec{x}_{P(1)},\ldots,\vec{x}_{P(2)} \rangle.$$
Here ##P## runs over all permutations ##S_N## of the indices of the positions of the particles. You can imagine that this becomes quite cumbersome if ##N## gets larger since the sum runs over ##N!## terms.
Now you can instead introduce bosonic annihilation and creation operators which create these basis states out of the "vacuum" state, which is defined as the situation that no particle is present. Then you define the creation operators recursively by
$$\hat{\psi}^{\dagger}(\vec{x}) |\vec{x}_1,\ldots,\vec{x}_N \rangle^{+}=|\vec{x},\vec{x}_1,\ldots,\vec{x}_N \rangle^{+}.$$
This makes a completely symmetrized product of position basis vectors for ##N## indisinguishable bosons a completely symmetrized product of positions basis vectors for ##N+1## indistinguishable particles. If you denote "the vacuum" with ##|\Omega \rangle## (which describes the situation that no particle is there), then this leads to
$$|\vec{x}_1,\ldots,\vec{x}_N \rangle^{+} = \hat{\psi}^{\dagger}(\vec{x}_1)\cdots \hat{\psi}^{\dagger}(\vec{x}_N) |\Omega \rangle.$$
The definition of the creation operators from the symmetrization of the bosonic basis states implies the commutation relation
$$[\hat{\psi}^{\dagger}(\vec{x}_1),\hat{\psi}^{\dagger}(\vec{x}_2)]=0,$$
so that with this definition the symmetrization is automatically guaranteed. That makes it much easier to deal with indistinguishable particles, even if you have a fixed number of states.
One can derive from this definition that the adjoint operator is an annihilation operator, particularly annihilating the vacuum state (i.e., it maps the vacuum state to the null vector).
It's also possible to define all kinds of operators of the first-quantization formalism to an equivalent operator in this second-quantization formalism. Now, any Hamiltonian of the first-quantization formalism keeps the particle number the same. You cannot even easily formulate the creation and annihilation of particles in the first-quantization formalism. This implies that if you express any model of the first-quantization formalism into its equivalent formulation in the second-quantization formalism your Hamiltonian consists of a sum of field-operator expressions where you always have as many creation and annihilation field operators in each term, i.e., applying the Hamiltonian to any completely symmetrized ##N##-particle basis vector doesn't lead into a space with a different particle number. So for this cases the 2nd-quantization formalism is indeed completely equivalent with the 1st-quantization formalism. It then only makes the task to always stay in the properly symmetrized (or for fermions antisymmetrized) many-particle spaces much easier.
