I am not sure if this is a duplicate. Clifford Taubes assert in his book Differential Geometry that we may view sections of vector bundles as homomorphisms from $M\times \mathbb{R}$ to $E$ such that for section $\sigma:M\rightarrow E$ we can define $f(m,r)\rightarrow r\sigma(m)$ and for homomorphism $f$ we can define section $\sigma(m)=f(m,1)$.
Given this construction as a context, I am not sure why he considered a section of $\otimes_{k}E^{*}$ defines a $k$-linear, fiber preserving map from $\oplus_{k}E$ to $M\times \mathbb{R}$. The fiber of $E^{*}$ at $m\in M$ is a vector space $V^{*}$ dual to $V$ which is the fiber at $m$ for $E$. If we view $\otimes_{k}V^{*}$ as the set of $k$-linear maps from $\oplus_{k} V$ to $\mathbb{R}$, then at the fibre level (for the same $m$ in $M$) a section of $M\rightarrow \otimes_{k}E^{*}$ implies a homomorphism $f:m\times \mathbb{R}\rightarrow \otimes_{k}V^{*}$. Now assume we have $f(m,r)=r\otimes^{k}_{i=1}g_{i}$ with $g_{i}$ certain maps in $V^{*}$. $f(m,r)$ can now be evaluated at $m$'s preimage $\oplus_{k}V$, and become a map $\oplus_{k}V\rightarrow \mathbb{R}$ as $f(m,r)(v_{1},.v_{k})=rg(v_{1})\times..\times g(v_{k})$
My confusion is this still does not look like a map from $\oplus V$ to $M\times \mathbb{R}$. Does this imply we may consider the map $(m,(v_{1},..v_{k}))\rightarrow (m, \prod g_{i}(v_{i}))?$Something must be wrong; so I venture to ask.