The general version you are looking for is the following: Suppose that$\Phi:V\to W$ is a vector bundle map between two vector bundles $V$ and $W$ over $M$. Suppose that we have connections $\nabla^V$ and $\nabla^W$, which are compatible with $\Phi$ in the sense that $\nabla^V_\xi \Phi(s)=\Phi(\nabla^W s)$ for all $s\in\Gamma(V)$ and for any vector field $\xi\in{\frak{X}}(M)$. Then $\Phi$ induces maps $\Phi_*:\Omega^k(M,V)\to\Omega^k(M,W)$ by acting on the values of forms. The general statement is that $\Phi_*$ is compatible with the covariant exterior derivatives, i.e. $d^{\nabla^W}(\Phi_*(\alpha))=\Phi_*(d^{\nabla^V}\alpha)$ for any $\alpha\in\Omega^k(M,V)$. The case you are looking at is $V=E\otimes E$ and $\nabla^V$ the induced connection from the metric connection $\nabla$ on $E$ and $W=M\times\mathbb R$ with the trivial connection (so the covariant exterior derivative becomes the exterior derivative) and $\Phi=tr_{\eta}$. The compatibility of $tr_{\eta}$ with the two connections essentially is the definition of a metric connection and it is verified your question.

Now for the proof of the general version, you just need the "global formula" for the covariant exterior derivative: Take vector fields $\xi_0,\dots,\xi_k\in{\frak X}(M)$ and a form $\beta\in\Omega^k(M,W)$. Then you can write $(d^{\nabla^W}\beta)(\xi_0,\dots,\xi_k)$ as $$ \sum_i(-1)^i\nabla^W_{\xi_i}\beta(\xi_0,\dots,\widehat{\xi_i},\dots,\xi_k)+\sum_{i<j}(-1)^{i+j}\beta([\xi_i,\xi_j],\xi_0,\dots,\widehat{\xi_i},\dots,\widehat{\xi_j},\dots,\xi_k) $$ with the hats denoting omission. If you insert $\beta=\Phi_*\alpha$, then in the second sum you simply get $\beta(\dots)=\Phi(\alpha(\dots))$. Doing a similar replacement in the first sum, you then use compatibility of $\Phi$ with the connections as $$ \nabla^W_{\xi_i}\Phi(\alpha(\xi_0,\dots,\widehat{\xi_i},\dots,\xi_k))= \Phi(\nabla^V_{\xi_i}\alpha(\xi_0,\dots,\widehat{\xi_i},\dots,\xi_k)). $$ Hence you conclude that $(d^{\nabla^W}\Phi_*\alpha)(\xi_0,\dots,\xi_k)=\Phi(d^{\nabla^V}\alpha(\xi_0,\dots,\xi_k))$ which is exactly the claim $d^{\nabla^W}\Phi_*\alpha=\Phi_*d^{\nabla^V}\alpha$.

The general version you are looking for is the following: Suppose that$\Phi:V\to W$ is a vector bundle map between two vector bundles $V$ and $W$ over $M$. Suppose that we have connections $\nabla^V$ and $\nabla^W$, which are compatible with $\Phi$ in the sense that $\nabla^V_\xi \Phi(s)=\Phi(\nabla^W s)$ for all $s\in\Gamma(V)$ and for any vector field $\xi\in{\frak{X}}(M)$. Then $\Phi$ induces maps $\Phi_*:\Omega^k(M,V)\to\Omega^k(M,W)$ by acting on the values of forms. The general statement is that $\Phi_*$ is compatible with the covariant exterior derivatives, i.e. $d^{\nabla^W}(\Phi_*(\alpha))=\Phi_*(d^{\nabla^V}\alpha)$ for any $\alpha\in\Omega^k(M,V)$. The case you are looking at is $V=E\otimes E$ and $\nabla^V$ the induced connection from the metric connection $\nabla$ on $E$ and $W=M\times\mathbb R$ with the trivial connection (so the covariant exterior derivative becomes the exterior derivative) and $\Phi=tr_{\eta}$. The compatibility of $tr_{\eta}$ with the two connections essentially is the definition of a metric connection and it is verified your question.

Now for the proof of the general version, you just need the "global formula" for the covariant exterior derivative: Take vector fields $\xi_0,\dots,\xi_k\in{\frak X}(M)$ and a form $\beta\in\Omega^k(M,W)$. Then you can write $(d^{\nabla^W}\beta)(\xi_0,\dots,\xi_k)$ as $$ \sum_i(-1)^i\nabla^W_{\xi_i}\beta(\xi_0,\dots,\widehat{\xi_i},\dots,\xi_k)+\sum_{i<j}(-1)^{i+j}\beta([\xi_i,\xi_j],\xi_0,\dots,\widehat{\xi_i},\dots,\widehat{\xi_j},\dots,\xi_k) $$ with the hats denoting omission. If you insert $\beta=\Phi_*\alpha$, then in the second sum you simply get $\beta(\dots)=\Phi(\alpha(\dots))$. Doing a similar replacement in the first sum, you then use compatibility of $\Phi$ with the connections as $$ \nabla^W_{\xi_i}\Phi(\alpha(\xi_0,\dots,\widehat{\xi_i},\dots,\xi_k))= \Phi(\nabla^V_{\xi_i}\alpha(\xi_0,\dots,\widehat{\xi_i},\dots,\xi_k)). $$ Hence you conclude that $(d^{\nabla^W}\Phi_*\alpha)(\xi_0,\dots,\xi_k)=\Phi(d^{\nabla^V}\alpha(\xi_0,\dots,\xi_k))$ which is exactly the claim $d^{\nabla^W}\Phi_*\alpha=\Phi_*d^{\nabla^V}\alpha$.

Edit: As mentioned in the comment, the compatibility condition should read as $\nabla^W_\xi \Phi(s)=\Phi(\nabla^V_\xi s)$ for all $\xi$ and $s$.