Complementary to ACuriousMind's answer (in particular, the crucial observation that what particle physics experiments measure are scattering amplitudes, hence the experimental relevance of the quantum fields stems primarily from the LSZ formula), and in the spirit of Robin Ekman's comment, you can also think of quantum fields as building bloc operators, from which to construct actual observables on a Fock space.
For a free, massive, complex scalar field $\Psi(x) = \Psi_{\text{part}}(x) + \Psi^{\dagger}_{\text{anti-part}}(x)$ on the Fock space $\mathcal{F} = \mathcal{F}_{\text{part}} ⊗ \mathcal{F}_{\text{anti-part}}$, this works as follow:
One-particle subspace
Let $\mathcal{H}_{\text{1 part}}$ be the subspace of the Fock space $\mathcal{F}_{\text{part}}$ spanned by states containing exactly 1 particle. A pseudo-basis of it can be labeled by impulsions $p$ on the mass-shell, with the normalization (using Weinberg's conventions):
$$
\left\langle p \middle| p' \right\rangle = \delta^{(3)}(\vec{p} - \vec{p}^{\prime})
$$
Let $A$ be some Hermitian operator on $\mathcal{H}_{\text{1 part}}$ (ie. a 1-particle observable) and, for simplicity, assume that $A$ has discrete spectrum, so we have an orthonormal basis of $\mathcal{H}_{\text{1 part}}$ made of eigenvectors:
$$
\left| \psi_k \right\rangle = \int d^{(3)}\vec{p}\; \tilde{\psi}_k(\vec{p}) \left| p \right\rangle
$$
with corresponding eigenvalues $\lambda_k$.
The integral kernel of $A$ is then:
$$
\tilde{A}(\vec{p},\vec{p}^{\prime}) := \sqrt{2 E(\vec{p}) 2 E(\vec{p}^{\prime})} \sum_k \tilde{\psi}_k(\vec{p}) \tilde{\psi}^*_k(\vec{p}^{\prime})
$$
or, in position representation (on the $t=0$ time slice):
$$
A(\vec{x},\vec{x}^{\prime}) := \int \frac{d^{(3)}\vec{p} \, d^{(3)}\vec{p}^{\prime}}{(2\pi)^3} e^{i(\vec{p}\cdot\vec{x} - \vec{p}^{\prime}\cdot\vec{x}^{\prime})} \tilde{A}(\vec{p},\vec{p}^{\prime})
$$
Creation/annihilation operators
A basis of the Fock space $\mathcal{F}_{\text{part}}$ built on $\mathcal{H}_{\text{1 part}}$ can be given in terms of occupation numbers over the basis $\big( \left| \psi_k \right\rangle \big)_k$:
$$
\left| \left( N_k \right)_k \right\rangle :=
\text{normalized symmetrization of }
\left| \psi_1 \right\rangle^{(1)}
\otimes \dots \otimes
\left| \psi_1 \right\rangle^{(N_1)}
\otimes \dots \otimes
\left| \psi_K \right\rangle^{(N_K)}
$$
Switching from the standard impulsion pseudo-basis into this $A$-adapted basis, one can check that the operators:
$$
a_{\text{part},k} := \int d^{(3)}\vec{p}\; \tilde{\psi}^*_k(\vec{p}) a_{\text{part}}(p) \;\&\; a^{\dagger}_{\text{part},k} := \int d^{(3)}\vec{p}\; \tilde{\psi}_k(\vec{p}) a^{\dagger}_{\text{part}}(p)
$$
annihilate, resp. create, a particle in the state $\left| \psi_k \right\rangle$.
Second quantization of $A$
The particle part of the quantum complex scalar field is given as:
$$
\Psi_{\text{part}}(x) := \int \frac{d^{(3)} \vec{p}}{(2\pi)^{3/2} \sqrt{2E(\vec{p})}} e^{i p \cdot x} a_{\text{part}}(p)
$$
so, putting everything together, we get:
$$\boxed{
\hat{A} := \int_{t=0} d^{(3)}\vec{x}\, d^{(3)}\vec{x}^{\prime}\; \Psi^{\dagger}_{\text{part}}(\vec{x},0)\, A(\vec{x},\vec{x}^{\prime})\, \Psi_{\text{part}}(\vec{x}^{\prime},0) = \sum_k λ_k \hat{N}_{\text{part},k}
}$$
where $\hat{N}_{\text{part},k} := a^{\dagger}_{\text{part},k} a_{\text{part},k}$ measures the number of particles in the state $\left| \psi_k \right\rangle$.
For example, for a region $U$ on the $t=0$ time slice,
$$
\int_{t=0, \vec{x} \in U} d^{(3)}\vec{x}\; \Psi^{\dagger}_{\text{part}}(\vec{x},0) \Psi_{\text{part}}(\vec{x},0)
$$
measures the number of particles in $U$ (the corresponding 1-particle operator $A$ has spectrum $\{0,1\}$ with the $(λ=1)$-eigenspace being spanned by wave-functions supported on $U$ only).
Assorted remarks
Of course, the $t=0$ slice is not distinguished: since Poincaré-invariance is built-in, you can switch to any space-like slice of Minkowski space (and, assuming I got all normalization factors right, this should go smoothly; no guarantee that I did, though...).
The "half-field" $\Psi_{\text{part}}(x)$ can be itself be reconstructed from $\Psi(x), \partial_t \Psi(x)$ (this is more easily seen in Fourier-transform). In a similar way, the underlying $\Psi_{\text{part}}(x)$ can be reconstructed from a real scalar field $\Phi(x) = \Psi_{\text{part}}(x) + \Psi^{\dagger}_{\text{part}}(x)$.
I am fairly confident that this reasoning can be generalized to other kinds of free fields, by carefully using the appropriate representation intertwinner to convert between spin numbers and field components (even if the field is fermionic, the observables constructed in this way, being quadratic in the field, are bosonic, so should be physically ok).
As mentioned by ACuriousMind, for interacting fields, this Fock picture is only valid asymptotically.
