My lecture notes define a connection on a vector bundle $\pi:E\rightarrow{M}$ to be an $\mathbb{R}$-linear map: \begin{equation} \nabla:\Gamma(E)\rightarrow\Gamma(T^*M\otimes{E}) \end{equation}
satisfying the Leibniz rule $\nabla(fs)=\mathrm{d}f\otimes{s}+f\nabla{s}$ for all $f\in{C^{\infty}(M)}$, $s\in\Gamma(E)$.
Then given a vector field $X$ on $M$, the covariant derivative along $X$ is the map \begin{equation} \nabla_X:\Gamma(E)\rightarrow\Gamma(E) \end{equation} such that for $s\in\Gamma(E)$, $\nabla_Xs=\mathrm{Tr}(X\otimes\nabla{s})$, that is, the contraction of the first two components of $X\otimes\nabla{s}\in{TM}\otimes{T^*M}\otimes{E}$ (I think it's okay to omit the $\Gamma$'s here for simplicity, and that this isn't a mistake?)
However, I'm trying to reconcile this definition with what I'm finding in almost every other source, which is as follows: with the same set up as above, a connection is a map \begin{equation} \nabla:\Gamma(TM)\times\Gamma(E)\rightarrow\Gamma(E) \end{equation}
satisfying $C^\infty$-linearity in the first component, $\mathbb{R}$-linearity in the second component, and the product rule $\nabla_X(fs)=f\nabla_X{s}+(Xf)s$ for all $f\in{C^\infty(M)}$. In this case, the image $\nabla_Xs$ of $(X,s)$ under $\nabla$ is defined to be the covariant derivative of $s$ in the direction $X$. This definition seems considerably easier to work with, so I would like to make the link between them.
I imagine it might have something to do with the identification between $\mathrm{Hom}(TM,E)$ and $T^*M\otimes{E}$, but I haven't got further than that.
If we partially apply your second definition to a section $s \in \Gamma(E)$, we get a $C^\infty(M)$-linear map $\nabla s : \Gamma(TM) \to \Gamma(E)$. You should know the fact that $C^\infty(M)$-linear maps between sheaves of sections are always derived from bundle homomorphisms ("$C^\infty$-linear maps are tensors"); i.e. $$\operatorname{Hom}_{C^\infty(M)}(\Gamma(TM), \Gamma(E)) = \Gamma(\operatorname{Hom}_\mathbb R(TM,E)).$$ Thus $\nabla s$ is in fact a section of $\operatorname{Hom} (TM,E) = T^* M \otimes E$ as desired, and the assignment $s \mapsto \nabla s$ satisfies the Leibniz rule in both definitions.