Here is an excerpt of Spivak's Differential Geometry.
What I do not understand is why he believes $\sigma \cdot (\rho \cdot v) = (\rho\sigma) \cdot v$. Since $\sigma$ and $\rho$ are elements of the symmetric group, they are permutations and thus form a group action which permutes tuples - in being a group action, they must satisfy $\sigma \cdot (\rho \cdot v) = (\sigma\rho) \cdot v$. For the obviously non-abelian symmetric group, it's easy to find examples where these are not equal.
Of course I don't think the book has such a glaring error in it, but I just don't see how it could be true. If you "just did it", you'd have $$\sigma\cdot(\rho\cdot (v_1\ldots v_k)) = \sigma\cdot(v_{\rho(1)}\ldots v_{\rho(k)}) = (v_{\sigma(\rho(1))}\ldots v_{\sigma(\rho(k))})$$ Which is clearly not always the same as his result. He says stuff about how if the $v$'s have their indices running in some different order, but who cares? $\sigma\cdot(v_3, v_2, v_1,\ldots ) = (v_{\sigma(3)}, v_{\sigma(2)}, v_{\sigma(1)}, \ldots)$, right? What's the problem??
There are two ways to think about sequences $(v_1,v_2,\ldots,v_k)$ here. Your way is to treat them as bijective functions $f:\{v_1,\ldots,v_k\}\to \{1,2,\ldots,k\}$ in which case the symmetric group acts by left composition, meaning $(\sigma\cdot f)(v_i)=\sigma(f(v_i))$. In that case it is easy to see that $\sigma\cdot (\rho\cdot f)=(\sigma\rho)\cdot f$.
Spivak's way, however, is to treat the sequences as functions $g:\{1,2,\ldots,k\}\to \{v_1,\ldots,v_k\}$ where the symmetric group acts by right composition, meaning $(\sigma\cdot g)(i)=g(\sigma(i))$. You can verify that in that case we indeed have that $\sigma\cdot (\rho\cdot g)=(\rho\sigma)\cdot g$.