Regarding Surjectivity in Serre's FAC.

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I started my first attempt at reading Serre's FAC and have found myself stuck on the sufficiency of the condition he gives in Proposition II (statement and given proof below). I've been able to figure out the necessity and assuming the open cover $\{U_i\}$ exists with $t_i$ as stated, I believe I understand how to conclude sufficiency, but I cannot seem to make the construction work. I fear I am missing something obvious and/or and consumed too much coffee to think straight.

The notation used in the paper: $\mathscr{F}_U$ the abelian group associated to $U$, $\Gamma(U,\mathscr{F})$, the sections of $\mathscr{F}$ over $U$, $\phi_U^V$ the morphism of abelian groups $\mathscr{F}_V \to \mathscr{F}_U$ for $U\subset V$, and $\iota:\mathscr{F}_U \to \Gamma(U,\mathscr{F})$ the morphism mapping $t$ to the section $x\mapsto \phi_x^U(t)$.

Proposition II: Let $U$ be an open subset of $X$ and $\iota: \mathscr{F}_U \to \Gamma(U,\mathscr{F})$ be injective for all open $V\subset U$. Then $\iota: \mathscr{F}_U \to \Gamma(U,\mathscr{F})$ is surjective if and only if the following condition is met:

For all open coverings $\{U_i\}$ of $U$ and all systems $\{t_i\}$, $t_i \in \mathscr{F}_{U_i}$ such that $\phi_{U_i \cap U_j}^{U_i} (t_i) = \phi_{U_i \cap U_j}^{U_j} (t_j)$ for all pairs $(i,j)$, there is some $t\in \mathscr{F}_U$ such that $\phi_{U_i}^U(t)=t_i$ for all $i$.

For a proof of sufficiency, the following is given:

The condition is sufficient: If $s$ is a section of $\mathscr{F}$ over $U$, there exists an open covering $\{U_i\}$ of $U$ and elements $t_i \in \mathscr{F}_{U_i}$ such that $\iota(t_i)$ coincides with the restriction of $s$ to $U_i$. It follows that the elements $\phi_{U_i \cap U_j}^{U_i}(t_i)$ and $\phi_{U_i \cap U_j}^{U_j}(t_j)$ define the same section over $U_i \cap U_j$ so by the assumption $\iota$ they are equal. If $t \in F_U$ satisfies $\phi_{U_i}^U(t)=t_i$, $\iota(t)$ coincides with $s$ over each $U_i$, thus over $U$.

Any help is appreciated.