Nobody really knows exactly what a quantum field is now because there is no rigorous mathematical formulation which is fully general, or even widely applicable. For example, the Standard Model of particle physics lacks such a foundation. But ignoring that for the moment, I think what would clear up your confusion most easily to have a look at the "Schrodinger picture" version of quantum field theory, as described in e.g. Hatfield, "Quantum Field Theory of Point Particles and Strings".
I am not going to reproduce all of that here but in short here is how you go from QM to QFT (for a free scalar field):
The wave function $\psi(x)$ becomes a wave functional $\Psi[\phi(\vec{x}),t]$, which gives the amplitude for observing a particular configuration of the entire field $\phi(\vec{x})$ at time $t$. You can also use a fourier basis and then the wave functional looks like $\Psi[\hat{\phi}(\vec{k})]$. (Note, this is not the $\hat{\Psi}$ from your question; I discuss that below).
I am not going to write the Lagrangian etc, but the result is that for a free field one gets a simple harmonic oscillator for each $\vec{k}$ mode. In other words there are operators $a_k$ and $a^\dagger_k$ which lower and raise the oscillator level (=occupation number) of that mode. The wave functional can then be written as $\Psi[n_\vec{k},t]$, i.e., the amplitude to find occupation numbers ${n_\vec{k}}$ at time $t$.
In QM one has the $X$ operator which multiplies the wave function by $x$. The corresponding operator in QFT is exactly the quantum field $\phi(\vec{x_0})$: it acts by multiplying $\Psi[\phi(\vec{x}),t]$. I wrote $\vec{x_0}$ to emphasize that you're only multiplying by one value, the value of $\phi$ at the given point $\vec{x_0}$.
So the quantum field is like the $X$ operator of QM, except there is a different one for every point of space. Now recall how for the oscillator, $X$ can be expressed in terms of $a,a^\dagger$. It is similar with fields: if you convert to the occupation basis $n_\vec{k}$ then the operation of multiplying by $\phi(\vec{x})$ gets expressed in terms of the $a_\vec{k}$, $a^\dagger_\vec{k}$. This leads (almost) to the integral form that you cited in your question.
I say "almost" because the integral as you wrote it shows the Heisenberg picture form of the operator. In the QM oscillator the analog is $X(t) \equiv \exp(iHt)X(0)\exp(-iHt)$. If you do this for the $a,a^\dagger$ operators then they will get time dependence $\exp(\pm i\omega t)$, which is what you see in your integral expression.
The other object that you showed, $\hat{\Psi}(x)$, is just the destruction-only part of $\phi(\vec{x})$, and it is written in a generic basis of wavefunctions $\psi_\nu$ rather than the fourier $\vec{k}$ basis. It has no time dependence because it is in the Schrodinger picture: time dependence is in the wave functional. This "destruction-only" part of the field isn't very useful in a relativistic theory because it can't be used to build Lorentz-invariant expressions; however, it's still a valid operator which "destroys a particle at position $x$".
So to summarize:
The two operators $\hat{\Psi}$,$\phi$ that you wrote are not the same. The first is the destruction-only piece of the field operator, written in Schrodinger picture. The second is the full field operator (as that term is used in relativistic QFT), written in Heisenberg picture.
