Note that $\theta$ being the difference of two connection one-forms, it is also horizontal: for every $\xi\in\mathfrak{g}$, $\theta(X_\xi)=0$, where $X_\xi$ is the fundamental vector field of the $G$-action of $P$ (i.e. $\theta$ vanishes on any section of the vertical bundle of $P$).
Therefore $\theta$ is basic, and thus determines a $1$-form $\bar\theta$ on $M$ with values in the adjoint bundle $ad(P)$ (that is the vector bundle associated with the adjoint representation).
Now I guess that your structure group $G$ comes with a morphism $G\to GL(r)$, where $r$ is the rank of $E$. This induces a morphism of representations $\mathfrak{g}\to End(\mathbb{R}^r)$. Passing to associated vector bundles, you get a bundle map $ad(P)\to End(E)$ (recall that $E$ is the vector bundle associated with the representation $\mathbb{R}^r$), and thus a linear map $\mu:\Omega^1(M,ad(P))\to \Omega^1(M,End(E))$.
The result you're looking after is $\mu(\bar\theta)=T$.
Remark: note that $\bar\theta(X)=\theta(X^h)$, and $\mu(\alpha)(X)(s)=\alpha(X)\cdot s$. Hence $\mu(\bar\theta)=T$ precisely becomes $\theta(X^h)\cdot s=T(X)(s)$.