The integral $$\int\limits_{U \times M \times \overline{U}} f(um\bar u) \delta_P(m)^{-1} d\bar u dm du = \int\limits_{P \times \overline{U}} f(p \bar{u}) d \bar u d_lp $$ is integration with respect to the product measure $d_lp d \bar{u}$ on the open dense set $P \times \overline{U} \cong P \overline{U}$ of $G$.  One has to check that the Haar measure on $G$ restricts to a Haar measure on the group $P \times \overline{U}$, even though $(p, \bar u) \mapsto p \bar u$ is not a group homomorphism. 

Therefore $\int\limits_{P \times \overline{U}} f(p \bar u) d \bar ud_lp$ is a scalar multiple of $\int\limits_G f(g) dg$, and to find the scalar, it suffices to compute the integral over $f = \operatorname{Char} K$.  The measure $dg$ is chosen so that $\int\limits_G f(g)dg = 1$.  

We extend $\delta_P$ to a continuous function on $G$ by making it trivial on $K$, so that $$\gamma(P) = \int\limits_{\overline{U}} \delta_P( \bar{u}) d \bar u$$  For each $\bar{u} \in \overline{U}$, write $\bar{u} = p_u k_u$ for $p_u \in P$ and $k_u \in K$ (nonuniquely), so that $\delta_P(\bar{u}) = \delta_P(p_u)$.  Now for $f = \operatorname{Char} K$,

$$\int\limits_{\overline{U}} \int\limits_P f(p \bar{u})d\bar{u} d_lp = \int\limits_{\overline{U}} \int\limits_P f(p p_u k_u) d_lp d \bar{u}$$

$$ = \int\limits_{\overline{U}} \int\limits_P f(pk_u) \delta_P(\bar{u}) d_lp d \bar{u} = \int\limits_{\overline{U}} \int\limits_{P \cap K} \delta_P(\bar{u}) d_lp d \bar{u} = \gamma(P)$$

since $d_lp(P \cap K) = 1$.  This gives the second equality in (2) and also guarantees the convergence of the integral defining $\gamma(P)$, since the Radon measure of the open compact set $K \cap P \overline{U}$ is finite and positive.