Hartshorne Proposition III.9.8, understanding associated points and extensions From Hartshorne III
Prop 9.8 Let Y be a regular scheme of dimension 1, let P $\in Y$ be a closed point, and let X $\subset \mathbf P_{Y-P}^n$ be a closed subscheme which is flat over Y - P. Then there exists a unique closed subscheme $\bar X \subset \mathbf P_Y^n$, flat over Y, whose restriction to $\mathbf P_{Y-P}^n$ is X.
The proof consists of taking $\bar X$ to be the set-theoretic closure of X. My trouble is in understanding why the associated points of $\bar X$ are the same as the associated points of X, and why any other extension would have some associated points mapping to P. How can other extensions look like?
By the way, how should I actually think about associated points? The only visual example that I have in mind is the "cross". Let A := k[x,y]/(xy), then the points (x), (y) and (x,y), only the last of which is closed, are all associated points because every element in the corresponding prime ideals is a zero-divisor. Does this mean that I should think of associated points as points that have some intersection with "another irreducible component"?
 A: In the ring $A=k[X,Y]/(X\cdot Y)=k[x,y]$  the prime ideals $(x),(y)$ are indeed associated, since they are minimal.
The  maximal ideal $(x,y)$ however is not associated since in a reduced  noetherian ring the converse of the above  is true: any associated prime is a minimal prime.    
Ravi Vakil in his magnificent algebraic geometry course   has a wonderful explanation of why and how the associated points of a scheme are its most important points.
Since it would be grotesque for me to try to improve upon his explanations, I will just direct you to page 164 of these notes.
A: I address the two main claims in the proof of Proposition III.9.8 in Hartshorne raised in the question.
Claim 1. The associated points of $X$ are the same as the associated points of $\bar{X}$.
This follows by understanding the nature of the scheme-theoretic image of a morphism of locally Noetherian schemes. On the level of affine schemes the scheme-theoretic image of the morphism $f: \operatorname{Spec} B \rightarrow \operatorname{Spec} A$ induced by some ring homomorphism $\phi: A \rightarrow B$ is (not surprisingly) $\operatorname{Spec} A / \ker \phi$. An important fact relating scheme-theoretic image and associated primes is the following. Let $f: \operatorname{Spec} A_g \rightarrow \operatorname{Spec} A$ be the open embedding corresponding to the canonical map $\phi:A \rightarrow A_g$ for some $g \in A$. The scheme-theoretic image of $f$ is $\operatorname{Spec} A / \ker \phi$ and $$\ker \phi = (0): g^{\infty} = \{a \in A: \, g^n a = 0, \text{for some $n \ge 0$} \}$$ In the above $(0): g^{\infty}$ is the saturation of the zero ideal $(0)$ in $A$ by $g$ and it can be computed from the primary decomposition of $(0)$. Indeed, it is the intersection of those primary components of $(0)$ which do not contain any power of $g$. It follows immediately that the associated primes of $A_g$ and $A/(0): g^{\infty}$ are in bijection. A similar fact is true more generally:
Theorem 1 (Exercise 8.3.D in Vakil's November 2017 notes). Let $f:W \rightarrow Z$ be a locally closed embedding into a locally Noetherian scheme. Then the associated points of the scheme-theoretic image of $f$ coincide with those of $W$.
Now Claim 1 follows from Theorem 1 and the fact that $X \rightarrow \mathbb{P}_{Y\setminus P}^n \rightarrow \mathbb{P}_{Y}^n$ is a locally closed embedding.
Claim 2. Any other extension of $X$ besides its scheme-theoretic closure would have associated points mapping to $P$.
I give a sketch of the affine local picture. Let $Y=\operatorname{Spec} A$ with $A$ a regular Noetherian domain of dimension $1$. With $P \in Y$ a closed point we can write $U = Y \setminus \{P\} = \cup_{i \in \mathcal{T}} D(g_i)$ where the $g_i$'s are ideal generators of $P$. Let $I,J$ be two ideals of $A[x]:=A[x_1,\dots,x_n]$ such that for every $i$ we have that $A_{g_i}[x] / I A_{g_i}[x]$ and $A_{g_i}[x] / J A_{g_i}[x]$ are isomorphic as $A_{g_i}[x]$-algebras. Moreover suppose that $\operatorname{Spec} A_{g_i}[x]/J A_{g_i}[x] \rightarrow \operatorname{Spec} A_{g_i}$ is flat. Then by Proposition III.9.7 every associated point of $\operatorname{Spec} A_{g_i}[x]/J A_{g_i}[x]$ maps to the generic point of $\operatorname{Spec} A_{g_i}$. Suppose that $\operatorname{Spec} A[x]/J$ is the scheme-theoretic image of $\operatorname{Spec} A_{g_i}[x]/J A_{g_i}[x] \rightarrow \operatorname{Spec} A[x]/J$, i.e. the map $A[x]/J \rightarrow A_{g_i}[x]/J A_{g_i}[x]$ is injective. On the other hand suppose that $I \neq J$. Then the map $A[x]/I \rightarrow A_{g_i}[x]/I A_{g_i}[x]$ is not injective. Hence $\bar{g}_i$ is a zero-divisor of $A[x]/I$ so that it lies in some $Q \in \operatorname{Ass} A[x]/I$. The contraction of $Q$ to $A$ must contain $g_i$ so that it does not map to the generic point $\eta$ of $Y$. In fact, $Q$ must contain every $g_i$ because otherwise it must map to $\eta$ by hypothesis. Since $\dim A=1$ we must have that $Q \cap A = P$.
