I think this works:

Consider a topological space consisting of 4 points $A$, $B$, $C$, $D$, where the topology is given by open sets $ABC$, $BCD$, $B$, $C$, $ABCD$, $\emptyset$.

Then let the presheaf $\mathcal{F}$ be given by:
$$\mathcal{F}(ABC)=\mathbb{Z}$$
$$\mathcal{F}(BCD)=\mathbb{Z}$$
$$\mathcal{F}(BC)=\mathbb{Z}$$
$$\mathcal{F}(ABCD)=\mathbb{Z}$$
$$\mathcal{F}(B)=\mathbb{Z}/2\mathbb{Z}$$
$$\mathcal{F}(C)=\mathbb{Z}/2\mathbb{Z}$$
$$\mathcal{F}(\emptyset)=0$$

where all restrictions are what you expect (identity in the case of $\mathbb{Z} \to \mathbb{Z}$ and canonical surjection in the case $\mathbb{Z} \to \mathbb{Z}/2 \mathbb{Z}$).

Then if we we get $\mathcal{F}^+$ is given by:

$$\mathcal{F}^+(ABC)=\mathbb{Z}$$
$$\mathcal{F}^+ (BCD)=\mathbb{Z}$$
$$\mathcal{F}^+ (BC)= \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$$
$$\mathcal{F}^+ (ABCD)=\mathbb{Z}$$
$$\mathcal{F}^+ (B)= \mathbb{Z}/2\mathbb{Z} $$
$$\mathcal{F}^+ (C)=\mathbb{Z}/2\mathbb{Z}$$
$$\mathcal{F}^+ (\emptyset)=0$$

where the map from $\mathcal{F}^+ (BCD)$ to $\mathcal{F}^+ (BC)$ is given by taking the canonical surjection on both copies, and other restrictions are obvious. Then note that if we take 1 over $BCD$ and 3 over $ABC$, these two are compatible over $BC$ but they do not patch.

The key point is that being compatible over a refinement is not the same thing as being compatible. That is, the way the plus construction works is by taking $F^+$ of a space to be some direct limit over open covers of guys on the covers which are compatible on intersections. If we had said instead take direct limit over open covers of guys on the covers which compatible on some refinement of the intersection, then applying just once probably works.

So in our example, 1 and 3, over $ABC$ and $BCD$, in our original presheaf were compatible on a refinement of $BC$ but not on $BC$.