For any (local) observable $\mathcal{O}$, its expectation value is defined as
$$ \langle \mathcal{O} \rangle = \frac{1}{Z}\int \mathcal{D}\Phi\mathcal{O}\mathrm{e}^{-S_E[\Phi]}$$
where
$$ Z = \int \mathcal{D}\Phi\mathrm{e}^{-S_E[\Phi]}$$
is indeed called the partition function. My writing $S_E$ is intended to show that this is for a Euclidean theory (i.e. a theory on a Riemannian manifold), for a Minkowskian theory (i.e. a theory on a Lorentzian manifold), one would have to add some $\mathrm{i}$ by Wick rotation. I have used $\mathcal{D}\Phi$ instead of $[\mathrm{d}\Phi]$ to denote the measure on the space of field configurations, both notations are widely used. It is important to note that, in most cases, this measure cannot be rigorously constructed and is only defined in some regularization procedure, but naively you can think of it as any kind of integration over the space of all possible field configurations $\Phi : \Sigma \rightarrow \mathbb{R}$, if $\Sigma$ is your spacetime manifold.
Now, your intuition about this thing are basically correct - $\mathrm{e}^{-S_E[\Phi]}$ is indeed a Boltzmann-like weighting factor, that, in the QFT case, weighs different field configurations $\Phi$. It is obvious that this factor will be maximal at minimal $S_E[\Phi]$, so the classically realized field configuration $\Phi_0$ for which $S_E[\Phi_0]$ is a minimum of the action contributes the most to this integral. The expectation value of the observable is indeed defined as an exact analogon to classical statististical mechanics.
You write something about a transformation, and from the context of Ward identities, I am assuming that you are talking about a gauge transformation, which should be a symmetry of the theory. Explicitly, the transformation is infintesimally given by
$$ \Phi \mapsto (1 + \mathrm{i}\omega_a(x)T^a)\Phi \equiv \Phi' \tag{1}$$
where the $T^a$ are generators of the symmetry group.
Being a symmetry of the theory means that no observable may change its value under the transformation. We denote quantities after transformation by primes. Looking at some observable $\mathcal{O}$ (your $X$), we must have that
$$ \langle \mathcal{O} \rangle = \frac{1}{Z}\int \mathcal{D}\Phi\mathcal{O}\mathrm{e}^{-S_E[\Phi]} \overset{!}{=} \frac{1}{Z'}\int \mathcal{D}\Phi' (\mathcal{O} + \delta\mathcal{O})\mathrm{e}^{-S_E[\Phi]-\int\mathrm{d}^d x \partial_\mu j^\mu_a \omega_a(x)}$$
We now assume that our symmetry is non-anomalous, i.e. $\mathcal{D}\Phi' = \mathcal{D}\Phi$. Then, for the equality to hold, we have:
$$\int \mathcal{D}\Phi\mathcal{O}\mathrm{e}^{-S_E[\Phi]} \overset{!}{=} \int \mathcal{D}\Phi (\mathcal{O} + \delta\mathcal{O})\mathrm{e}^{-S_E[\Phi]-\int\mathrm{d}^d x \partial_\mu j^\mu_a \omega_a(x)}$$
If you now expand $\mathrm{e}^{-\int\mathrm{d}^d x \partial_\mu j^\mu_a \omega_a(x)}$ to first order in $\omega_a$, you will get $\langle \mathcal{O} \rangle$ on both sides of the equation, which cancels, and the remaining terms are precisely the Ward identities, i.e. the first equation you asked about.
The second equation follows from doing the explicit transformation on $\mathcal{O}$: (Assuming, as you said, that the $G_a$ are the generators $T^a$ of the transformation): Look again at $(1)$. Since you said that $\mathcal{O}$ is a collection of fields, it is
$$ \mathcal{O} = \prod_{i=1}^n \Phi(x_i)$$
for some $n$. Now, carry out $(1)$ on every field $\Phi$ and keep only the terms at most first order in $\omega_a$. This is exactly your $\delta \mathcal{O}$.
(This is "exact" since we consider an infinitesimal gauge transformation, anyways. There are rigorous foundations in Lie theory for this "throwing away" of higher order terms. Nevertheless, it is quite important to carry these tricks the physicist likes so much oneself, since the answer coming out is correct.)
