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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.

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Your question is hard to read, but basically your last line is correct: for a decomposable element g=product of functions g_i the associated map sends v_1, , v_k to product of g_i(v_i). – Max Sep 9 '12 at 1:54
@Max: Where is this unclear? I thought I have provided all detail required. – Bombyx mori Sep 9 '12 at 2:00
I think mathematically it is fine; I just personally find your style of writing difficult (I am no English teacher, but I think that there are at least two run-on sentences, a couple of omitted "that", and a "Does this implies". Plus some minor gripes. Almost all of this is grammatically correct, but makes it harder for me to understand what you are saying.) Apart from that, what you wrote is correct. Of the last two maps in the post the very last one is the correct one, and the preceding one is not. You seem to have answered your own question. – Max Sep 9 '12 at 2:58
@Max: Thanks. I understand. It is embarrassing that I am still suffering from standard English grammar mistakes. Thanks a lot for your response. – Bombyx mori Sep 9 '12 at 3:25
I was also going to claim this was hard to read, before @Max's comments, but I would have said it was because you use many variable names without defining them. The notation is standard, but the best practice is that an MSE question should be as readable to those who aren't experienced in a field as possible. I also would say $E^{*\otimes^k}$ or $(E^{*\otimes})^k$ for $\otimes_kE^*$, since the latter looks like a tensor over the field $k$. – Kevin Carlson Sep 9 '12 at 6:16

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