$\newcommand{\de}{\delta}
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\newcommand{\R}{\mathbb{R}}
\newcommand{\E}{\operatorname{\mathsf E}} 
\newcommand{\PP}{\operatorname{\mathsf P}}$

The answer is no, in general it is not possible to express the measure $(\mu * \nu)_h$ in terms of the families $(\mu_h)_{h \in H}$ and $(\nu_h)_{h \in H}$ only. 

E.g., suppose that $G$ is the additive group $\R^2$, $N=\R\times\{0\}$, measures $\mu,\hat\mu,\nu$ are probability measures, 

$\bullet\quad$ $\mu$ attaches masses $\frac28,\frac18,\frac58$ to points $(0,0),(0,1),(1,0)$, respectively;
  
$\bullet\quad$ $\hat\mu$ attaches masses $\frac24,\frac14,\frac14$ to points $(0,0),(0,1),(1,0)$, respectively; 

$\bullet\quad$ $\nu$ attaches masses $\frac14,\frac14,\frac24$ to points $(0,0),(0,1),(1,0)$, respectively.  

Then $\mu_h=\hat\mu_h$ for $p_* \mu$-almost all $h \in H$ and, equivalently, for $p_* \hat\mu$-almost all $h \in H$. 

However, 
$$(\mu * \nu)_1(\{0\})=\frac9{16}\ne\frac58=(\hat\mu * \nu)_1(\{0\}).$$