I am trying to understand the line bundle $O(1)$ over $\Bbb P^1_k$ and why
$$ O(1)^{\otimes n} = O(n)$$ from Vakil's notes, p398, line 6.
His explanation is rather long, so I took a screen shot.
I don't understand Vakil's explanation of why we have the equality because of the transition functions.
On
I was trying to understand this claim as well and I found it helpful to have done Vakil 13.1.D first, as per @KReiser's suggestion. We can view the situation via transition functions
We claim that for $n \geq 0$, we have the same description for $\mathscr{O}_\mathbb{P}$-modules
$$ \mathscr{O}(n) =\mathscr{O}(1)^{\otimes n} $$
reasoning: see that $\mathscr{O}(n)$ is a locally free sheaf wrt $\mathscr{O}_\mathbb{P}$ of rank $1$, since it is given on the standard affine cover by $\mathscr{O}_\mathbb{P}|_{U_i}$ itself. If $t_{ij}^{(k)}$ gives the transition function for the $k^{th}$ copy of $\mathscr{O}(1)$, then, the transition function for $\mathscr{O}(1)^{\otimes n}$ is given by the multilinear map $\bigotimes_{k}^n t_{ij}^{(k)}$ (see Vakil 13.1.D) To see what this is in detail, recall the formula for the canonical isomorphism $R \otimes_R M \cong M$ :
$$ r \otimes_R m \mapsto r \cdot m \\ m \mapsto 1 \otimes_R m $$
Since we know that $t^{(k)}_{ij}(-) $is $ x_{j/i}\cdot (-)$, the transition function for $\mathscr{O}(1)^{\otimes n}$ on $V \subseteq U_i \cap U_j$, going from $U_i$ to $U_j$, is exactly
$$ \begin{align*} p(x_{i/j}) &\leftrightarrow 1 \otimes_{\mathscr{O}_\mathbb{P}} 1 \otimes_{\mathscr{O}_\mathbb{P}(V)} \cdots \otimes_{\mathscr{O}_\mathbb{P}(V)} p(x_{i/j}) \\ &\mapsto x_{j/i} \otimes_{\mathscr{O}_\mathbb{P}(V)} x_{j/i} \otimes_{\mathscr{O}_\mathbb{P}(V)}\cdots \otimes_{\mathscr{O}_\mathbb{P}(V)} p(\frac 1 {x_{j/i}}) \\ & =1 \otimes_{\mathscr{O}_\mathbb{P}(V)} 1 \otimes_{\mathscr{O}_\mathbb{P}(V)} \cdots \otimes_{\mathscr{O}_\mathbb{P}(V)} x_{j/i} ^n \cdot p(\frac 1 {x_{j/i}}) \\ & \leftrightarrow x_{j/i} ^n \cdot p(\frac 1 {x_{j/i}}) \end{align*} $$
This is exactly the transition function for $\mathscr{O}(n)$, so we have shown that
$$ \mathscr{O}(n) = \mathscr{O}(1)^{\otimes n} $$
When $n < 0$, we claim that there is an isomorphism of $\mathscr{O}_\mathbb{P}$-modules
$$ \mathscr{O}(n) \cong \mathscr{O}(-n)^\vee $$
reasoning: once again, $\mathscr{O}(n)$ is still a locally free sheaf wrt $\mathscr{O}_\mathbb{P}$ of rank 1, for the exact same reasons as before. To show that $\mathscr{O}(n) \cong \mathscr{O}(-n)^\vee$, it suffice (see Vakil 13.1.D) to show that $\mathscr{O}(n)$ is the inverse of $\mathscr{O}(-n)$ in the Picard group, i.e. we want to show that $\mathscr{O}(-n) \otimes_{\mathscr{O}_\mathbb{P}} \mathscr{O}(n) \cong \mathscr{O}_\mathbb{P}$.
But as we have done above, observe that the transition functions for the tensor product $\mathscr{O}(-n) \otimes_{\mathscr{O}_\mathbb{P}} \mathscr{O}(n)$ is given by multilinear maps $t_{ij}^{(-n)} \otimes t_{ij}^{(n)}$ which on the intersection, sends $p(x_{i/j}) \in k[x_{i/j}]$ to
$$ \begin{align*} p(x_{i/j}) &\leftrightarrow 1\otimes_{\mathscr{O}_\mathbb{P}(V)} p(x_{i/j}) \\ &\mapsto x_{j/i} ^{-n}\otimes_{\mathscr{O}_\mathbb{P}(V)} x_{j/i}^n \cdot p(\frac 1 {x_{j/i}}) \\ & = 1 \otimes_{\mathscr{O}_\mathbb{P}} x_{j/i}^{-n}\cdot x_{j/i} ^n \cdot p(\frac 1 {x_{j/i}}) \\ & \leftrightarrow p(\frac 1 {x_{j/i}}) \end{align*} $$
This is the transition function for $\mathscr{O}_\mathbb{P}$, which proves our claim.
The whole point here is that if we take two invertible sheaves $\scr L$ and $\scr M$ on $X$ and an open cover $\{U_i\}$ of $X$ such that $\mathscr L$ and $\scr M$ are both trivializable, so that $\scr L$ and $\scr M$ are given by transition functions (satisfying the cocyle conditions) over the intersections, say by $f_{ij}\in\mathscr O_X(U_{ij})^\times$ and $g_{ij}\in\mathscr O_X(U_{ij})^\times$ respectively (I am writing $U_{ij}:=U_i\cap U_j$), then $\mathscr L\otimes\scr M$ is also trivializable on the cover $\{U_i\}$ and the associated transition functions for the tensor product on this cover are exactly given by $f_{ij}g_{ij}$ over $U_{ij}$.
How to see this? Well let's recall quickly where the "transition functions" are coming from; for each $i$ we have isomorphisms $\varphi_i:\mathscr L|_{U_i}\to\mathscr O_X|_{U_i}$, and then over $U_{ij}$ we have an isomorphism $$\mathscr O_X|_{U_{ij}}\overset{\varphi_i^{-1}|_{U_{ij}}}\longrightarrow\mathscr L|_{U_{ij}}\overset{\varphi_j|_{U_{ij}}}\longrightarrow\mathscr O_X|_{U_{ij}}$$
but any such isomorphism is determined by a global section of $\mathscr O_X|_{U_{ij}}$, given by the image of $1\in\mathscr O_{X}(U_{ij})$ under the map on sections of $U_{ij}$, and this is exactly the element $f_{ij}$ we are calling the transition function (sorry for the drawn out details if this is all obvious to you); similarly we have $g_{ij}$ coming from $\psi_j\circ\psi_i^{-1}$ (actually the restrictions to $U_{ij}$ as we wrote more carefully above), where $\psi_i:\mathscr M|_{U_i}\to\mathscr O_X|_{U_i}$ are some local trivializations for $\mathscr M$.
Now notice we have a sequence of isomorphisms
$$(\mathscr L\otimes\mathscr M)|_{U_i}\longrightarrow\mathscr L|_{U_i}\otimes\mathscr M|_{U_i}\overset{\varphi_i\otimes\psi_i}\longrightarrow\mathscr O_X|_{U_i}\otimes\mathscr O_X|_{U_i}\longrightarrow\mathscr O_X|_{U_i}$$ (the composition of which we will call $\rho_i$) where the last map is just multiplication, and this gives us explicit trivializations of $\mathscr L\otimes\mathscr M$ on the cover $\{U_i\}$. Now that we know more precisely what the transitions exactly are from our discussion above, let's compute them with respect to our chosen trivializations $\rho_i$; the map $\rho_j\circ\rho_i^{-1}$ on sections over $U_{ij}$ is explicitly the composition
$$\mathscr O_X(U_{ij})\to\mathscr O_X(U_{ij})\otimes\mathscr O_X(U_{ij})\overset{\varphi_i^{-1}\otimes\psi_i^{-1}}\longrightarrow\mathscr L(U_{ij})\otimes\mathscr M(U_{ij})\overset{\varphi_j\otimes\psi_j}\longrightarrow\mathscr O_X(U_{ij})\otimes\mathscr O_X(U_{ij})\to\mathscr O_X(U_{ij})$$
and then if we chase down where $1\in\mathscr O_X(U_{ij})$ goes under this map we find
$$1\mapsto 1\otimes 1\mapsto \varphi_i^{-1}(1)\otimes\psi_i^{-1}(1)\mapsto(\varphi_j\circ\varphi_i^{-1})(1)\otimes(\psi_j\circ\psi_i^{-1})(1)=f_{ij}\otimes g_{ij}\mapsto f_{ij}g_{ij}$$
proving $f_{ij}g_{ij}$ indeed are the transition functions of the tensor product over this cover.
Hope this helps; when I was learning this stuff I was furious there were tons of details like this nobody was being precise about, so I tried to be as explicit as I could.