Proof.
Note that in (2) the morphism $x : \mathop{\mathrm{Spec}}(K) \to X$ defines a $K$-rational point of $X_ z$, hence the statement makes sense. Moreover, the condition in (2) is independent of the choice of $\mathop{\mathrm{Spec}}(K) \to X$ in the equivalence class of $x$ (details omitted; this will also follow from the arguments below because the other conditions do not depend on this choice). Also note that we can always choose a diagram as in (3) by: first choosing a scheme $W$ and a surjective étale morphism $W \to Z$, then choosing a scheme $V$ and a surjective étale morphism $V \to W \times _ Z Y$, and finally choosing a scheme $U$ and a surjective étale morphism $U \to V \times _ Y X$. Having made these choices we set $U \to W$ equal to the composition $U \to V \to W$ and we can pick a point $u \in U$ mapping to $x$ because the morphism $U \to X$ is surjective.
Suppose given both a diagram as in (3) and a geometric point $\overline{x} : \mathop{\mathrm{Spec}}(k) \to X$ as in (1). By Properties of Spaces, Lemma 66.19.4 we can choose a geometric point $\overline{u} : \mathop{\mathrm{Spec}}(k) \to U$ lying over $u$ such that $\overline{x} = a \circ \overline{u}$. Denote $\overline{v} : \mathop{\mathrm{Spec}}(k) \to V$ and $\overline{w} : \mathop{\mathrm{Spec}}(k) \to W$ the induced geometric points of $V$ and $W$. In this setting we know that $\mathcal{O}_{X, \overline{x}} = \mathcal{O}_{U, u}^{sh}$ and similarly for $Y$ and $Z$, see Properties of Spaces, Lemma 66.22.1. In the same vein we have
\[ \mathcal{F}_{\overline{x}} = (a^*\mathcal{F})_ u \otimes _{\mathcal{O}_{U, u}} \mathcal{O}_{U, u}^{sh} \]
see Properties of Spaces, Lemma 66.29.4. Note that the stalk of $\mathcal{F}_ w$ at $u$ is given by
\[ (\mathcal{F}_ w)_ u = (a^*\mathcal{F})_ u/\mathfrak m_ w(a^*\mathcal{F})_ u \]
and the local ring of $V_ w$ at $v$ is given by
\[ \mathcal{O}_{V_ w, v} = \mathcal{O}_{V, v}/\mathfrak m_ w\mathcal{O}_{V, v}. \]
Since $\mathfrak m_{\overline{z}} = \mathfrak m_ w \mathcal{O}_{Z, \overline{z}} = \mathfrak m_ w \mathcal{O}_{W, w}^{sh}$ we see that
\begin{align*} \mathcal{F}_{\overline{x}}/ \mathfrak m_{\overline{z}}\mathcal{F}_{\overline{x}} & = (a^*\mathcal{F})_ u \otimes _{\mathcal{O}_{U, u}} \mathcal{O}_{X, \overline{x}}/ \mathfrak m_{\overline{z}}\mathcal{O}_{X, \overline{x}} \\ & = (\mathcal{F}_ w)_ u \otimes _{\mathcal{O}_{U_ w, u}} \mathcal{O}_{U, u}^{sh}/\mathfrak m_ w\mathcal{O}_{U, u}^{sh} \\ & = (\mathcal{F}_ w)_ u \otimes _{\mathcal{O}_{U_ w, u}} \mathcal{O}_{U_ w, \overline{u}}^{sh} \\ & = (\mathcal{F}_ w)_{\overline{u}} \end{align*}
the penultimate equality by Algebra, Lemma 10.156.4 and the last equality by Properties of Spaces, Lemma 66.29.4. The same arguments applied to the structure sheaves of $V$ and $Y$ show that
\[ \mathcal{O}_{V_ w, \overline{v}}^{sh} = \mathcal{O}_{V, v}^{sh}/\mathfrak m_ w \mathcal{O}_{V, v}^{sh} = \mathcal{O}_{Y, \overline{y}}/ \mathfrak m_{\overline{z}}\mathcal{O}_{Y, \overline{y}}. \]
OK, and now we can use Morphisms of Spaces, Lemma 67.31.1 to see that (1) is equivalent to (3).
Finally we prove the equivalence of (2) and (3). To do this we pick a field extension $\tilde K$ of $K$ and a morphism $\tilde x : \mathop{\mathrm{Spec}}(\tilde K) \to U$ which lies over $u$ (this is possible because $u \times _{X, x} \mathop{\mathrm{Spec}}(K)$ is a nonempty scheme). Set $\tilde z : \mathop{\mathrm{Spec}}(\tilde K) \to U \to W$ be the composition. We obtain a commutative diagram
\[ \xymatrix{ & & & U_ w \times _ w \tilde z \ar[llld]_ a \ar[rr] \ar[dr] & & V_ w \times _ w \tilde z \ar[llld]_>>>>>>>b \ar[dl] \\ X_ z \ar[rr]_ f \ar[dr]_ g & & Y_ z \ar[dl]^ h & & \tilde z \ar[llld]_ c \\ & z } \]
where $z = \mathop{\mathrm{Spec}}(K)$ and $w = \mathop{\mathrm{Spec}}(\kappa (w))$. Now it is clear that $\mathcal{F}_ w$ and $\mathcal{F}_ z$ pull back to the same module on $U_ w \times _ w \tilde z$. This leads to a commutative diagram
\[ \xymatrix{ X_ z \ar[d] & U_ w \times _ w \tilde z \ar[l] \ar[d] \ar[r] & U_ w \ar[d] \\ Y_ z & V_ w \times _ w \tilde z \ar[l] \ar[r] & V_ w } \]
both of whose squares are cartesian and whose bottom horizontal arrows are flat: the lower left horizontal arrow is the composition of the morphism $Y \times _ Z \tilde z \to Y \times _ Z z = Y_ z$ (base change of a flat morphism), the étale morphism $V \times _ Z \tilde z \to Y \times _ Z \tilde z$, and the étale morphism $V \times _ W \tilde z \to V \times _ Z \tilde z$. Thus it follows from Morphisms of Spaces, Lemma 67.31.3 that
\[ \mathcal{F}_ z\text{ flat at }x\text{ over }Y_ z \Leftrightarrow \mathcal{F}|_{U_ w \times _ w \tilde z} \text{ flat at }\tilde x\text{ over }V_ w \times _ w \tilde z \Leftrightarrow \mathcal{F}_ w\text{ flat at }u\text{ over }V_ w \]
and we win.
$\square$
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