= Solution
A <connection on a vector bundle> is a real-linear map (complex-linear for a complex bundle)
$$
D:\Gamma(E)\longrightarrow\mathcal A^1(E),\qquad
D(fs)=df\otimes s+fDs.
$$
It extends to the <covariant exterior derivative> by $D(\eta\otimes s)=d\eta\otimes s+(-1)^k\eta\wedge Ds$ when $\eta$ has degree $k$. Its square is linear over smooth functions: in $D^2(fs)$ the two $df\wedge Ds$ terms have opposite signs and $d^2f=0$. The <curvature form of a connection> is therefore the endomorphism-valued two-form
$$
\boxed{\Theta=D^2\in\mathcal A^2(\operatorname{End}E).}
$$
In a local frame with $D=d+\theta$, it is $\Theta=d\theta+\theta\wedge\theta$.
Define the <tensor product connection> on decomposable sections by
$$
\boxed{D(s\otimes t)=D_1s\otimes t+s\otimes D_2t.}
$$
The <Leibniz rule> for $D_1,D_2$ makes this well-defined over smooth functions and makes it a <connection on a vector bundle>. Applying its <covariant exterior derivative> again gives
$$
D^2(s\otimes t)=D_1^2s\otimes t-D_1s\wedge D_2t
+D_1s\wedge D_2t+s\otimes D_2^2t.
$$
The mixed terms cancel by the graded sign. Thus \b[curvature adds on the two tensor factors]:
$$
\boxed{\Theta=\Theta_1\otimes I+I\otimes\Theta_2.}
$$
Equivalently, the mixed terms in $(\theta_1\otimes I+I\otimes\theta_2)\wedge(\theta_1\otimes I+I\otimes\theta_2)$ cancel because operators on the different factors commute. This also proves the <curvature of a tensor product connection> formula in local frames.
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