I'm reading this note in chapter 6 which discusses $ {\cal N} = 2 $ supermultiplets, though the discussion is quite general and applies to $ {\cal N} = 1 $ fields in the adjoint representation. I summarize the discussion given in the paper here with a slightly simplified situation and notation for clarity. Suppose we have some field, $ \Phi $, in the adjoint representation of a gauge group $SU(N)$. The Kahler term for the field is

\begin{equation}\int d ^2 \theta \Phi ^\dagger e ^{ 2 g V ^a T ^a} \Phi =g D ^a \phi ^\dagger T ^a \phi + ...\end{equation}

and the kinetic gauge term gives,

\begin{equation}W^\alpha _a W_\alpha^a=\frac{1}{2} D ^a D ^a+...\end{equation}

The paper then goes on to integrate out the $D^a$ fields which gives,

\begin{equation}D^a =\Phi ^\dagger T ^a \Phi\end{equation}

Then the authors go on to introduce more ${\cal N} =1$ field, $H$. The $H$ field Lagrangian takes the form,

\begin{align}{\cal L} &= \int d^4\theta H ^\dagger e^{2q V ^a t ^a} H+ \int d^2\theta W^\alpha _a W_\alpha^a + ...\\&= g D ^a \Phi ^\dagger T ^a \Phi + \frac{1}{2}D_aD_a...\end{align}

where $t_a$ are the generators in the representation of the $H$ fields.

So far I have no issues, however then do something strange. They go on to again integrate out the $D$ fields, but they do so independently of the fields introduced earlier in the adjoint representation. However, don't we have a single $\vec{D}$ field for every gauge group, not every representation? So how come we are able to consider the sectors independently of one another?