With your notations, the hermitian form $\theta_E$ on $T_X\otimes E$ defined by $\Theta_E$ is given in a somewhat more extrinsic way by
$$
\theta_E(v\otimes\sigma,v\otimes\sigma):=h(\Theta_E(v,\bar v)\cdot \sigma,\sigma),
$$
where

- $h$ is the hermitian metric on $E$,
- $v\in T_X$, so that $\bar v\in \overline{T_X}$ (or, if you want, $v\in T^{1,0}_X$ so that $\bar v\in T^{0,1}_X$),
- $\sigma\in E$.

Note that I gave the formula just for decomposable (rank one) tensors, but then you extend it to all tensors by sesquilinearity.

Now, let $(E,h_E)$ and $(F,h_F)$ be two holomorphic hermitian vector bundles. Take the product metric $h_{E\otimes F}=h_{E}\otimes h_F$ on their tensor product so that, as you said, the curvature of the corresponding Chern connection is given by
$$
\Theta(E\otimes F)=\Theta(E)\otimes\operatorname{Id}_F+\operatorname{Id}_E\otimes\Theta(F).
$$
Then, for $v\in T_X$, $\sigma\in E$ and $\tau\in F$, you get
$$
\begin{aligned}
\theta_{E\otimes F}(v\otimes\sigma\otimes\tau,v\otimes\sigma\otimes\tau) &=
h_{E\otimes F}(\Theta_{E\otimes F}(v,\bar v)\cdot(\sigma\otimes\tau),\sigma\otimes\tau) \\
&=h_{E\otimes F}((\Theta(E)(v,\bar v)\otimes\operatorname{Id}_F \\
&\qquad\qquad\qquad\qquad\qquad+\operatorname{Id}_E\otimes \Theta(F)(v,\bar v))\cdot(\sigma\otimes\tau),\sigma\otimes\tau) \\
&=h_{E\otimes F}((\Theta(E)(v,\bar v)\cdot\sigma)\otimes\tau+\sigma\otimes(\Theta(F)(v,\bar v)\cdot\tau),\sigma\otimes\tau) \\
&=h_{E\otimes F}((\Theta(E)(v,\bar v)\cdot\sigma)\otimes\tau,\sigma\otimes\tau) \\
&\qquad+h_{E\otimes F}(\sigma\otimes(\Theta(F)(v,\bar v)\cdot\tau),\sigma\otimes\tau) \\
&=h_E(\Theta(E)(v,\bar v)\cdot\sigma,\sigma)h_F(\tau,\tau) \\
&\qquad+h_E(\sigma,\sigma)h_F(\Theta(F)(v,\bar v)\cdot\tau,\tau) \\
&=\theta_E(v\otimes\sigma,v\otimes\sigma)h_F(\tau,\tau)+h_E(\sigma,\sigma)\theta_F(v\otimes\tau,v\otimes\tau).
\end{aligned}
$$
Thus, the decomposition you were looking for is perhaps the most natural one:
$$
\theta_{E\otimes F}=\theta_E\otimes h_F+h_E\otimes\theta_F.
$$

**EDIT:** Re-edited the answer coherently with this other answer.