If I understand you correctly you believe in the counter example and just want an intuitive reason. Here is how I think about it (I am open for remarks!):

Definition: Let $\mathscr A$ be a (small) category with arbitrary (small) limits. A presheaf $F$ on a (small) site $\mathscr C$ with values in $\mathscr A$ is a sheaf, if the diagrams $F(U)\rightarrow \prod_i F(U_i)\rightrightarrows \prod_{i,j}F(U_i\times_UU_j)$ are equalizers for all covering families $U_i\rightarrow U$ of all objects $U$ of $\mathscr C$.

Construction: Let $F$ be an $\mathscr A$-valued presheaf on a site $\mathscr C$ and $U\in \mathscr C$. Then for any covering family $\mathscr U=\{U_i\rightarrow U\}$, define $F_U(\mathscr U)$ to be the equalizer of the sequence $F_U(\mathscr U)\rightarrow \prod_i F(U_i)\rightrightarrows \prod_{i,j}F(U_i\times_UU_j).$ Further define $F^+(X)$ as the colimit $F^+(X)=colim_{\mathscr U}F_X(\mathscr U)$.

And here is what happens: We do the whole construction for each $U$ seperately. And this is precisely the dangerous part: Look at the set $U=ABC$ in Sherry's example. Here, the set $BC$ is part of a covering and thus has influence on $F^+(U)$. But at the same time we apply our construction to the set $BC$, which doesn't even see the element $A$. So while we are trying to adjust our sheaf on $ABC$, so that it fits with $BC$, we change the sheaf on $BC$ independently. In the end more sections are compatible than before, but we didn't include their glueings yet. Morally the reason for that phenomenon is (non-)separatedness. By Definition we just have precisely one glueing for every compatible family in $F^+$. That means on the one hand that $F^+$ is automatically separated, but it means also that if $F$ is not separated then the different glueings for the same compatible family become "identified". In our example we have the elements $1$ and $3$ in $F(BC)$, which give the same compatible family in $F(C)$ and $F(B)$. In $F^+(BC)$ we have just one element for every family in $F(C)$ and $F(B)$.So $F^+(BC)$ looks quite different than $F(BC)$.

If I understand you correctly you believe in the counter example and just want an intuitive reason. Here is how I think about it (I am open for remarks!):

Definition: Let $\mathscr A$ be a (small) category with arbitrary (small) limits. A presheaf $F$ on a (small) site $\mathscr C$ with values in $\mathscr A$ is a sheaf, if the diagrams $F(U)\rightarrow \prod_i F(U_i)\rightrightarrows \prod_{i,j}F(U_i\times_UU_j)$ are equalizers for all covering families $U_i\rightarrow U$ of all objects $U$ of $\mathscr C$.

Construction: Let $F$ be an $\mathscr A$-valued presheaf on a site $\mathscr C$ and $U\in \mathscr C$. Then for any covering family $\mathscr U=\{U_i\rightarrow U\}$, define $F_U(\mathscr U)$ to be the equalizer of the sequence $F_U(\mathscr U)\rightarrow \prod_i F(U_i)\rightrightarrows \prod_{i,j}F(U_i\times_UU_j).$ Further define $F^+(U)$ as the colimit $F^+(U)=colim_{\mathscr U}F_U(\mathscr U)$.

And here is what happens: We do the whole construction for each $U$ seperately. And this is precisely the dangerous part: Look at the set $U=ABC$ in Sherry's example. Here, the set $BC$ is part of a covering and thus has influence on $F^+(U)$. But at the same time we apply our construction to the set $BC$, which doesn't even see the element $A$. So while we are trying to adjust our sheaf on $ABC$, so that it fits with $BC$, we change the sheaf on $BC$ independently. In the end more sections are compatible than before, but we didn't include their glueings yet. Morally the reason for that phenomenon is (non-)separatedness. By Definition we just have precisely one glueing for every compatible family in $F^+$. That means on the one hand that $F^+$ is automatically separated, but it means also that if $F$ is not separated then the different glueings for the same compatible family become "identified". In our example we have the elements $1$ and $3$ in $F(BC)$, which give the same compatible family in $F(C)$ and $F(B)$. In $F^+(BC)$ we have just one element for every family in $F(C)$ and $F(B)$.So $F^+(BC)$ looks quite different than $F(BC)$.