Let $k$ be a field and $k((x))$ be its field of formal laurent series. I'm trying to understand morphisms $f : Spec \ k((x)) \rightarrow \mathbb{P} = Proj \ k[T_0, T_1]$. For example, does specifying that $f(0)=p \in \mathbb{P}$ and $f^{\#}(T_1/T_0) = s \in k((x))$, where we assumed $T_0 \notin p$, specify a unique morphism?
My attempt:
Let's say we have the information above. We have to construct a sheaf morphism $f^{\#}: O_{\mathbb{P}} \rightarrow f_{*}O_{k((x))}$. Pick any distinguished open in $\mathbb{P}$, say $U_g$. Then $O_{\mathbb{P}} = (k[T_0,T_1]_g)_0$. However $O_{k((x))}(f^{-1}U_g) = 0$ if $g \in p$ and $k((x))$ if not. Let $f^{\#}_g : (k[T_0,T_1]_g)_0 \rightarrow O_{k((x))}(f^{-1}U_g)$.
So $g \in p \implies f^{\#}_g = 0$. Since $T_0 \notin p$, it makes sense to set $f^{\#}_{T_0}(T_1/T_0) = s$. From this, we can deduce for example that if $T_1 \notin p$, $f^{\#}_{T_1}(T_0/T_1) = s^{-1}$, since we can restrict to $U_{T_0T_1}$. Or if $T_1-T_0 \notin p$, then $f^{\#}_{T_1-T_0}T_0/(T_1-T_0) \rightarrow (s-1)^{-1} \in k((x)).$ In general, if $g \notin p$ is any homogenous polynomial, then $(k[T_0,T_1]_g)_0$ is a subring of $k(T_1/T_0)$ and so the image of elements of this ring under $f^{\#}_g$ are determined by the image of $T_1/T_0$, which we already know?
Is this enough to show this morphism is well defined and unique?
In general, if $K$ is a field, morphisms from $\text{Spec}(K)$ to any scheme $X$ correspond to the data of a point $x \in |X|$ together with a fixed embedding of fields from the residue field of $x$ to $K$ (note that, crucially, there need not be just one of these embeddings). This can be seen pretty quickly by just noticing that one can localise on $X$: since $|\text{Spec}(K)|$ is a one-point topological space, we have that any morphism $\text{Spec}(K) \to X$ will necessarily factor through any open subset $U \subset X$ containing its singleton image; if $U \cong \text{Spec}(A)$ if an affine such neighbourhood, then $\text{Spec}(K) \to \text{Spec}(A)$ is given by a ring homomorphism $A \to K$ which factors through the quotient $A/\mathfrak{p}_x$ where $\mathfrak{p}_x \subset A$ is the prime ideal corresponding to $x \in \text{Spec}(A) \implies$ since $K$ is a field, we get an embedding $\text{res}(x) := \text{Frac}(A/\mathfrak{p}_x) \hookrightarrow K$. The described procedure can of course be reversed.
Having said that (and I apologise if my discussion only repeated things which were already clear to you), there is definitely no unique way to embed the residue field of any point in $\mathbb{P}^1_k$ into $k((t))$, even if $k$ is algebraically closed. For instance, if you take the generic point $\eta \in \mathbb{P^1_k}$, its residue field is the field of rational functions $k(t)$, and the canonical inclusion $k(t) \hookrightarrow k((t))$ is not the only embedding: you can map $t \in k(t)$ to pretty much anything you like actually - e.g. $$ k(t) \to k((t)),$$
$$ a \in k \mapsto a \in k((t)), $$
$$ t \mapsto \sum\limits_{i=-n}^\infty\, a_i t^i $$ defines a perfectly valid ring homomorphism so long as $\sum\limits_{i=-n}^\infty\, a_i t^i$ is transcendental over $k$ (which occurs for instance if $a_{-n} \neq 0$).
I'm not sure this completely answers your question, but I hope it helps! :)