Linearising shallow-water equations about a base state 
Consider a shallow-water system with
  mean depth $H$, where the base state consists of the 
  flow $(u,v)=(u_{0},0)$, with a sloped water surface $\eta_{0}(x,y)
= - \gamma y$, where $u_{0}$ and $\gamma$ are constants. 
  Consider now small amplitude deviations $u'$ $v'$ $\eta'$ from this base
  state and assume for simplicity 
  that none of the deviations vary with $y$ (ie the partials of the primes
  wrt to $y$ are zero). With this 
  assumption linearise the shallow water system about the mean state.

For this question do we have to linearise the equations 
$u=u_{0}+u'$
$v= 0+v' $
$\eta = -\gamma y + \eta'$
if i am right here, i am unsure how to proceed. 
any pointers please?, A previous example I have 
used the momentum equation $\frac{DU'}{DT}+g\frac{\partial\eta'}{\partial x}-fv'=0$
and expanded it to $\frac{\partial u'}{\partial t}+u'\frac{\partial u'}{\partial x} +v' \frac{\partial u'}{\partial y} +g \frac{\partial\eta'}{\partial x}-fv'=0$
then neglected the middle two terms and dropped the primes
Im a little unsure of this process and help understanding would be 
appreciated. 
 A: This can only be answered denpending on what formulation of the shallow water equations you consider. Based on the formula you presented, I suspect you use the shallow water equations for constant density $\rho$ with  viscous effects but respected coriolis force $f$ in non-conservative form.
The notation for the different heights is not really consistent, I assume you apply the one used in this resource, which also deals with linearization around a certain state.
\begin{align} 
\partial_t \eta+ \partial_x \Big((\eta +h) u \Big) + \partial_y \Big( (\eta +h) v \Big) & = 0 \\
\frac{D u}{Dt} - fv + g \partial_x \eta & = 0 \\
\frac{D v}{Dt} - fu + g \partial_y \eta & =  0 \end{align}
Since you have three unknowns, you must have three equations at hand - that's why only the momentum equations do not suffice.
For $u_0 \neq u_0(x, y, t), \gamma \neq \gamma(x, y, t)$ you get for the presented linearizations:
\begin{align} 
\partial_t \eta'+ \partial_x \Big((-\gamma y + \eta' +h) (u_0 + u') \Big) + \partial_y \Big( (-\gamma y + \eta' +h) v' \Big) & = 0 \\
\partial_tu' + (u_0 + u') \partial_x u' + v' \partial_y u' - fv' + g \partial_x \eta' & =  0 \\
\partial_t v' + (u_0 + u') \partial_x v' + v' \partial_y v' - f(u_0 + u') + g (- \gamma \partial_y \eta') &=  0 \end{align}
To obtain truly linear equations, no products of $(\cdot)'$ quantities (and derivatives) may remain. They are typically neglected under the reasoning that for varaitions of order $\mathcal{O}(\Delta)$, variations of order $\mathcal{O}(\Delta^2)$ are negligible for $\Delta \ll 1$.
This gives
\begin{align} 
\partial_t \eta'+ \partial_x \Big(-\gamma y u_0  + \eta' u_0 + h u_0 - \gamma y u' + hu'  \Big) + \partial_y \Big( -\gamma yv' +h v' \Big) & = 0 \\
\partial_tu' + u_0 \partial_x u' - fv' + g \partial_x \eta' & =  0 \\
\partial_t v' + u_0 \partial_x v'  - f(u_0 + u') + g (- \gamma \partial_y \eta') &=  0 \end{align}
and after further simplification of the first equation you obtain a linear system of unknowns $\eta', u', v'$ with parameters $f, g, h, u_0, \gamma$.
\begin{align} 
\partial_t \eta'+ u_0 \partial_x\eta' - \gamma y \partial_x u' + h \partial_x u'  -\gamma ( v' + y \partial_y v') +h \partial_y v'& = 0 \\
\partial_tu' + u_0 \partial_x u' - fv' + g \partial_x \eta' & =  0 \\
\partial_t v' + u_0 \partial_x v'  - f(u_0 + u') + g (- \gamma \partial_y \eta') &=  0 \end{align}
