So this is a follow up investigation on invertible modules on $Spec(K)$. Now I wish to understand $$Mor(Spec K, \Bbb P^n_{\Bbb Z})$$ which I have learnt is
represented by the set of morphisms $$ \bigoplus^n_0 K \rightarrow K $$ this is given by set of maps $(e_0,\ldots, e_n)$ where $e_i \in K$ and some $e_j \not=0$. Up to $K$-isomoprhisms $\beta:K^{n+1} \rightarrow K^{n+1}$, where $\beta(v_i)=e_i$.
This is supposed to equal to, according to nlab,
the set of lines through the origin in $K^{n+1}$
Edit: Now what I don't see is how the equivalence $\beta$ is equivalent to scalar multiplication - shouldn't our relation in the first case be: $v \sim w$ iff exists $A \in GL(n+1)$, $Av=w$?
Your translation of the property in your link is flawed. What it written there implies that the morphisms $\operatorname{Spec}(K)\to \mathbb{P}^n$ correspond to $(n+1)$-uples of sections of $\mathcal{O}_K$ (because it is the only invertible module, as you identified) over $\operatorname{Spec}(K)$ which generate the module, so indeed elements $(e_0,\dots,e_n)\in K^{n+1}$ with not all $e_i$ equal to $0$, and $(e_0,\dots,e_n)$ gives the same element as $(f_0,\dots,f_n)$ if there is an automorphism of $\mathcal{O}_K$ that sends $e_i$ to $f_i$. Now such an automorphism is indeed a scalar multiplication.