Let $\mathcal{L}$ be an invertible $\mathcal{O}_U$-module. Observe that $\mathcal{L}$ maps to $0$ in $\text{Pic}(U_0)$ if and only if we can extend $\mathcal{L}$ to an invertible coherent triple $(\mathcal{L}, \mathcal{L}_0, \lambda)$ as in Section \ref{section-coherent-triples}. By Proposition \ref{proposition-hilbert-triple} the function $$ n \longmapsto \chi((\mathcal{L}, \mathcal{L}_0, \lambda)^{\otimes n}) $$ is a polynomial $P$. By Lemma \ref{lemma-nonnegative-chi-triple} we have $P(n) \geq 0$ for all $n \in \mathbf{Z}$ with equality if and only if $\mathcal{L}^{\otimes n}$ is trivial. In particular $P(0) = 0$ and $P$ is either identically zero and we win or $P$ has even degree $\geq 2$. \medskip\noindent Set $M = \Gamma(U, \mathcal{L})$ and $M_0 = \Gamma(X_0, \mathcal{L}_0) = \Gamma(U_0, \mathcal{L}_0)$. Then $M$ is a finite $A$-module of depth $\geq 2$ and $M_0 \cong A/fA$, see proof of Lemma \ref{lemma-nonnegative-chi-triple}. Note that $H^2_\mathfrak m(M)$ is finite $A$-module by Local Cohomology, Lemma \ref{local-cohomology-lemma-local-finiteness-for-finite-locally-free} and the fact that $H^i_\mathfrak m(A) = 0$ for $i = 0, 1, 2$ since $\text{depth}(A) \geq 3$. Consider the short exact sequence $$ 0 \to M/fM \to M_0 \to Q \to 0 $$ Lemma \ref{lemma-nonnegative-chi-triple} tells us $Q$ has finite length equal to $\chi(\mathcal{L}, \mathcal{L}_0, \lambda)$. We obtain $Q = H^1_\mathfrak m(M/fM)$ and $H^i_\mathfrak m(M/fM) = H^i_\mathfrak m(M_0) \cong H^i_\mathfrak m(A/fA)$ for $i > 1$ from the long exact sequence of local cohomology associated to the displayed short exact sequence. Consider the long exact sequence of local cohomology associated to the sequence $0 \to M \to M \to M/fM \to 0$. It starts with $$ 0 \to Q \to H^2_\mathfrak m(M) \to H^2_\mathfrak m(M) \to H^2_\mathfrak m(A/fA) $$ Using additivity of lengths we see that $\chi(\mathcal{L}, \mathcal{L}_0, \lambda)$ is equal to the length of the image of $H^2_\mathfrak m(M) \to H^2_\mathfrak m(A/fA)$. \medskip\noindent Let prove the lemma in a special case to elucidate the rest of the proof. Namely, assume for a moment that $H^2_\mathfrak m(A/fA)$ is a finite length module. Then we would have $P(1) \leq \text{length}_A H^2_\mathfrak m(A/fA)$. The exact same argument applied to $\mathcal{L}^{\otimes n}$ shows that $P(n) \leq \text{length}_A H^2_\mathfrak m(A/fA)$ for all $n$. Thus $P$ cannot have positive degree and we win. In the rest of the proof we will modify this argument to give a linear upper bound for $P(n)$ which suffices. \medskip\noindent Let us study the map $H^2_\mathfrak m(M) \to H^2_\mathfrak m(M_0) \cong H^2_\mathfrak m(A/fA)$. Choose a normalized dualizing complex $\omega_A^\bullet$ for $A$. By local duality (Dualizing Complexes, Lemma \ref{dualizing-lemma-special-case-local-duality}) this map is Matlis dual to the map $$ \text{Ext}^{-2}_A(M, \omega_A^\bullet) \longleftarrow \text{Ext}^{-2}_A(M_0, \omega_A^\bullet) $$ whose image therefore has the same (finite) length. The support (if nonempty) of the finite $A$-module $\text{Ext}^{-2}_A(M_0, \omega_A^\bullet)$ consists of $\mathfrak m$ and a finite number of primes $\mathfrak p_1, \ldots, \mathfrak p_r$ containing $f$ with $\dim(A/\mathfrak p_i) = 1$. Namely, by Local Cohomology, Lemma \ref{local-cohomology-lemma-sitting-in-degrees} the support is contained in the set of primes $\mathfrak p \subset A$ with $\text{depth}_{A_\mathfrak p}(M_{0, \mathfrak p}) + \dim(A/\mathfrak p) \leq 2$. Thus it suffices to show there is no prime $\mathfrak p$ containing $f$ with $\dim(A/\mathfrak p) = 2$ and $\text{depth}_{A_\mathfrak p}(M_{0, \mathfrak p}) = 0$. However, because $M_{0, \mathfrak p} \cong (A/fA)_\mathfrak p$ this would give $\text{depth}(A_\mathfrak p) = 1$ which contradicts assumption (4). Choose a section $t \in \Gamma(U, \mathcal{L}^{\otimes -1})$ which does not vanish in the points $\mathfrak p_1, \ldots, \mathfrak p_r$, see Properties, Lemma \ref{properties-lemma-quasi-affine-invertible-nonvanishing-section}. Multiplication by $t$ determines a map $t : M \to A$ which defines an isomorphism $M_{\mathfrak p_i} \to A_{\mathfrak p_i}$ for $i = 1, \ldots, r$. Via $M_0 \cong A/fA$ we may and do view $t \bmod f$ as an element $t_0 \in A/fA$. We conclude that there is a commutative diagram $$ \xymatrix{ \text{Ext}^{-2}_A(M, \omega_A^\bullet) & \text{Ext}^{-2}_A(M_0, \omega_A^\bullet) \ar[l] \\ \text{Ext}^{-2}_A(A, \omega_A^\bullet) \ar[u]^t & \text{Ext}^{-2}_A(A/fA, \omega_A^\bullet) \ar[l] \ar[u]_{t_0} } $$ It follows that the length of the image of the top horizontal map is at most the length of $\text{Ext}^{-2}_A(A/fA, \omega_A^\bullet)$ plus the length of the cokernel of $t_0$. \medskip\noindent However, if we replace $\mathcal{L}$ by $\mathcal{L}^n$ for $n > 1$, then we can use $$ t^n : M_n = \Gamma(U, \mathcal{L}^{\otimes n}) \longrightarrow \Gamma(U_0, \mathcal{L}_0^{\otimes n}) = M_{n, 0} $$ instead of $t$. This replaces $t_0 \in A/fA$ by its $n$th power. Thus the length of the image of the map $\text{Ext}^{-2}_A(M_n, \omega_A^\bullet) \leftarrow \text{Ext}^{-2}_A(M_{n, 0}, \omega_A^\bullet)$ is at most the length of $\text{Ext}^{-2}_A(A/fA, \omega_A^\bullet)$ plus the length of the cokernel of $$ t_0^n :  \text{Ext}^{-2}_A(A/fA, \omega_A^\bullet) \longrightarrow \text{Ext}^{-2}_A(A/fA, \omega_A^\bullet) $$ Since $\text{Ext}^{-2}_A(A/fA, \omega_A^\bullet)$ is a finite $A$-module with support of dimension $1$ as indicated above this length grows linearly in $n$ by Algebra, Lemma \ref{algebra-lemma-support-dimension-d}. \medskip\noindent To finish the proof we prove the final assertion. Assume $f \in \mathfrak m \subset A$ satisfies (1), (2), (3), $A$ is $(S_2)$, and $\dim(A) \geq 4$. Condition (1) implies $A$ is catenary, see Dualizing Complexes, Lemma \ref{dualizing-lemma-universally-catenary}. Then $\Spec(A)$ is equidimensional by Local Cohomology, Lemma \ref{local-cohomology-lemma-catenary-S2-equidimensional}. Thus $\dim(A_\mathfrak p) + \dim(A/\mathfrak p) \geq 4$ for every prime $\mathfrak p$ of $A$. Then $\text{depth}(A_\mathfrak p) \geq \min(2, \dim(A_\mathfrak p)) \geq \min(2, 4 - \dim(A/\mathfrak p))$ and hence (4) holds.