Proof.
Let us observe that given $V$ and $V \to Y$ as in (2) without any further assumption on $f$ we see that the morphism $X \times _ Y V \to V$ corresponds to a surjective morphism $B \to A$ in $\textit{WAdm}^{Noeth}$ by Formal Spaces, Lemma 87.29.5.
Assume (1). By Lemma 88.21.4 we see that $\text{Spf}(A) \to \text{Spf}(B)$ is rig-surjective. Let $I \subset B$ be an ideal of definition. Since $B$ is adic, $I^ m \subset B$ is an ideal of definition for all $m \geq 1$. If $I^ m J^ n \not= 0$ for all $n, m \geq 1$, then $IJ$ is not nilpotent, hence $V(IJ) \not= \mathop{\mathrm{Spec}}(B)$. Thus we can find a prime ideal $\mathfrak p \subset B$ with $\mathfrak p \not\in V(I) \cup V(J)$. Observe that $I(B/\mathfrak p) \not= B/\mathfrak p$ hence we can find a maximal ideal $\mathfrak p + I \subset \mathfrak m \subset B$. By Algebra, Lemma 10.119.13 we can find a discrete valuation ring $R$ and an injective local ring homomorphism $(B/\mathfrak p)_\mathfrak m \to R$. Clearly, the ring map $B \to R$ cannot factor through $A = B/J$. According to Lemma 88.21.11 this contradicts the fact that $\text{Spf}(A) \to \text{Spf}(B)$ is rig-surjective. Hence for some $n, m$ we do have $I^ n J^ m = 0$ which shows that (2) holds.
Assume (2). By Lemma 88.21.8 it suffices to show that $\text{Spf}(A) \to \text{Spf}(B)$ is rig-surjective. Pick an ideal of definition $I \subset B$ and an integer $n$ such that $I J^ n = 0$. Consider a ring map $B \to R$ where $R$ is a discrete valuation ring and the image of $I$ is nonzero. Since $R$ is a domain, we conclude the image of $J$ in $R$ is zero. Hence $B \to R$ factors through the surjection $B \to A$ and we are done by definition of rig-surjective morphisms.
$\square$
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