Bayes Correlated Equilibrium: obedience
To understand the "Obedience" inequality, notice that the player is "integrating out" everything that she is uncertain about, this is why the sum runs over the actions of other players, $a_{-i}$, the types of other players, $t_{-i}$, and the state of the world, $\theta$. 

The first term simply gives the probability that the mediator observed $\theta$. 

For the second term, note that it is proportional to the conditional probability of the other players being of type $t_{-i}$ when you are of type $t_i$ since $\pi(t_{-i}|t_i,\theta)=\frac{\pi(t_i,t_{-i}|\theta)}{\sum_{t_{-i}}\pi(t_i,t_{-i}|\theta)}$. 

Similarly for the third term, you probably expected there to see the conditional probability $\sigma(a_{-i}|a_i,t_i,t_{-i},\theta)$, but this term is proportional to $\sigma(a_i, a_{-i}|t_i,t_{-i},\theta)$.

Technically speaking the agent should multiply his payoff by the probability that others play $a_{-i}$, are of type $t_{-i}$ and the state is $\theta$ conditional on being told to play $a_i$ when being of type $t_i$, so the LHS should read:

$$\sum_{a_{-i}\in A_{-i}, t_{-i}\in T_{-i}, \theta\in\Theta}u_i((a_i,a_{-i}),\theta) \frac{\sigma((a_i,a_{-i})|(t_i,t_{-i}),\theta)\pi((t_i,t_{-i})|\theta)\psi(\theta)}{\sum_{a'_{-i}\in A_{-i}, t'_{-i}\in T_{-i}, \theta'\in\Theta}\sigma((a_i,a'_{-i})|(t_i,t'_{-i}),\theta')\pi((t_i,t'_{-i})|\theta')\psi(\theta')}$$

And the RHS will be identical except for replacing $a_i$ with $\tilde a_i$ in the utility function. Clearly, the denominator is constant over the sum and so can be simplified. 

Note: In previous versions on the paper the authors were more explicit about this derivation.