= Solution
Let the <Eulerian time average> commute with space and time derivatives and let the statistically steady mean have no time derivative. Define $U=\overline u$, $V=\overline v$, $\hat u=u-U$ and $\hat v=v-V$, so $\overline{\hat u}=\overline{\hat v}=0$. Averaging instantaneous <incompressibility> gives
$$
\boxed{U_x+V_y=0,}
$$
and subtracting gives $\hat u_x+\hat v_y=0$ for the fluctuations. Start from the instantaneous streamwise <Navier-Stokes equation>
$$
u_t+u u_x+v u_y=-\frac1\rho p_x+\nu(u_{xx}+u_{yy}).
$$
Use <incompressibility> to write $u u_x+v u_y=\partial_x(u^2)+\partial_y(uv)$. The averaged products are $\overline{u^2}=U^2+\overline{\hat u^2}$ and $\overline{uv}=UV+\overline{\hat u\hat v}$. The mean conservative terms reduce back to $UU_x+VU_y$ because $U_x+V_y=0$. The <Reynolds-averaged momentum equation> is consequently
$$
\boxed{UU_x+VU_y=-\frac1\rho\bar p_x+\nu U_{yy}-\partial_y\overline{\hat u\hat v}-\partial_x\overline{\hat u^2}+\nu U_{xx}.}
$$
The signs use the positive covariance convention for the <Reynolds stress>. In particular, both quadratic averages involve the fluctuations, not products of the mean $U$ with a fluctuation; the converted TeX loses the overbar/hat structure here.
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