Page 270, Lee's Introduction to Smooth Manifolds, 1st edition, Proposition 11.9:
Suppose $F: M \rightarrow N$ and $G:N \rightarrow P$ are smooth maps, $\sigma \in \mathcal{T}^k(N)$, $\tau \in \mathcal(T)^l(N)$, and $f \in C^\infty(N)$.
(Here $\mathcal{T}^k(N)$ is the vector space of covariant k-tensor fields over $N$, so $\sigma: N \rightarrow T^k(N)$ is a smooth section from $N$ into the k-covariant tensor bundle of $N$).
The first property stated is bit confusing, and the details are left as an exercise. It states that:
$F^*(f \sigma) = ((fF) F^* \sigma)$.
The next property says that $F^*(\sigma \otimes \tau) = F^*(\sigma) \otimes F^*(\tau)$, and that by identifying $f \otimes \sigma$ with $f \sigma$, that the first property is a special case of the second (this identification makes sense as we can regard $f$ as a 0-form).
Anyway, This was confusing me.
I know that $F^*(\sigma)(X_1,....X_k)= \sigma(F_*X_1,....F_*X_k)$
And I feel like $F^*(f \sigma) = ((fF) F^* \sigma)$, doesn't make sense, as $F^* \sigma$ already sends vectors in $M$ to $\mathbb{R}$, so what exactly is $fF$ doing?
I would appreciate any insight what so ever. Thank you!!
Since $F^*(f\sigma)$ is a $k$-covariant tensor field over $M$, to understand what it does you have to first evaluate it on a point $p\in M$, thus getting a $k$-linear map $(T_pM)^k\to\mathbb{R}$, then evaluate this linear map on $k$ vectors $X_1,\dots,X_k\in T_pM$. We get:
$$F^*(f\otimes \sigma)_p(X_1,\dots,X_k)=(F^*f\otimes F^*\sigma)_p(X_1,\dots,X_k)=(F^*f)_p\,\times[(F^*\sigma)_p(X_1,\dots,X_k)].$$
This last equality is important: when you have a $l$-covariant tensor $\omega$ and a $m$-covariant tensor $\eta$, then by definition $\omega\otimes\eta$ is a $l+m$ covariant one defined by $$(\omega\otimes\eta)(X_1,\dots,X_l,X_{l+1},\dots,X_{l+m}):=\omega(X_1,\dots,X_l)\times \eta(X_l+1,\dots,X_{l+m}),$$
so $\omega$ takes the first $l$ vectors and $\eta$ the last $m$ ones. In our case, since $F^*f$ is still a $0$-tensor, it will take $0$ vectors, and $(F^*f)_p$ is simply the scalar $(f\circ F)(p)$. Finally, we get
$$F^*(f\otimes \sigma)_p(X_1,\dots,X_k)=(f\circ F)(p)\,\times\sigma_p(F_{*,p}(X_1),\dots,F_{*,p}(X_k)).$$
So $F^*f$ acts only as a scalar multiplication. I hope this answers your question.