Let $\eta_j(i)$ denote the $i$th element of vector $\eta_j$ with $j =1,2$.
By definition, we have
$\text{cov}(\eta_1,\eta_2) = \left( \begin{array}{cc} cov(\eta_1(1),\eta_2(1)) & cov(\eta_1(1),\eta_2(2)) \\ cov(\eta_1(2),\eta_2(1)) & cov(\eta_1(2),\eta_2(2)) \end{array} \right),$
where direct substitution gives
$\text{cov}(\eta_1,\eta_2) = \left( \begin{array}{cc} cov(\xi_1+\xi_2,\xi_1) & cov(\xi_1+\xi_2,\xi_1-\xi_2) \\ cov(\xi_2,\xi_1) & cov(\xi_2,\xi_1-\xi_2) \end{array} \right).$
From here you apply the definition of the covariance operator for two scalar-valued random variables and that of independence. You should get
$\text{cov}(\eta_1,\eta_2) = \left( \begin{array}{cc} 1 & -1 \\ 0 & 2 \end{array} \right).$