Eigenstates of a bosonic field operator

Even though related questions are discussed here and here, I am still confused about the eigenstates of the field operator of a bosonic field $$ \hat{\phi}(\vec{x},t=0)|\phi\rangle=\phi(\vec{x})|\phi\rangle $$ Does this mean that all states of the QFT are related to a field configuration in space? The states are said to fulfill the completeness relation $$ \int d\phi(\vec{x})\,|\phi\rangle\langle\phi|=1. $$ The measure here means that we integrate over all field configurations? Or does it mean that we integrate over all values a field can take at position $\vec{x}$? It must be the first case since a state can not already be specified by just the eigenvalue with respect to the field operator at one point, right? This would be in agreement with $$ \langle\phi_a|\phi_b\rangle=\prod_\vec{x}\delta(\phi_a(\vec{x})-\phi_b(\vec{x})). $$ So shouldn't one rather write $$ \prod_\vec{x}\int d\phi(\vec{x})\,|\phi\rangle\langle\phi|=1. $$

How would one take the trace of an operator? $$ \text{tr}\,\hat{\mathcal{O}}=\int d\phi(\vec{x}) \langle\phi|\mathcal{O}|\phi\rangle $$ or $$ \text{tr}\,\hat{\mathcal{O}}=\prod_\vec{x}\int d\phi(\vec{x}) \langle\phi|\mathcal{O}|\phi\rangle $$ or even something different?

The formulas I used are from Kapusta "Finite Temperature Field Theory Principles and Applications", p. 12+13.