Baby example of a covariant derivative - mistake in Taubes' Differential Geometry?

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I'm reading Taubes' Differential Geometry, Section 11: Covariant derivatives and connections.

The introduction defines a covariant derivative as an $\Bbb R$-linear map $$\nabla:C^\infty(M,E)\to C^\infty(M,E\otimes T^*M)$$ satisfying the Leibniz rule $$\nabla (fs)=f\nabla(s)+s\otimes \operatorname{d}\!f$$ where $f\in C^\infty(M,\Bbb R)$. Here $E$ is a real vector bundle over a smooth manifold $M$, with fiber $V$.

The first example given is built by patching up derivatives defined on charts: if $s:x\mapsto (x,v(x))$ is a section of the trivial bundle $U\times V\to U$, then the derivative is defined as $ds: x\mapsto (x,\operatorname{d}\!v|_x)$ (which I believe should be written as $x\otimes \operatorname{d}\!v|_x$ but anyway).

This local definition is then pulled back via the chart map and patched up over $M$ thanks to a partition of unity.

It all makes sense to me, except for the fact that $\operatorname{d}\!v|_x$ does not lie in $T^*U$ as it should, but in $\operatorname{Hom}(TU,TV)$ and therefore the final result lies not in $C^\infty(M,E\otimes T^*M)$ but rather in $C^\infty(M,E\otimes \operatorname{Hom}(TM,TE))$.

Is there a trick here that I'm not seeing, or is this just an example of something slightly more general than what is announced?

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Since $V$ is a vector space, you just use the standard view of $dv|_x$ as a linear map $T_xM\to V$ i.e. as an element of $T^*_xM\otimes V$. The latter space is exactly the fiber over $x$ of $T^*M\otimes E$ when $E$ is the trivial bundle $U\times V$.