If $f(x, y)$ is a formal group law then so is $g(f(g^{-1}(x), g^{-1}(y))$ where $g$ is an invertible (under composition) formal power series. This suggests a strategy for writing down $n$-buds, namely pick a polynomial $g$, a group law $f$, and a polynomial approximation $h$ to $g^{-1}$ and then compute $g(f(h(x), h(y))$. If $h$ agrees with $g^{-1}$ modulo degree $n+1$ terms then this gives an $n$-bud. 

For $g$ let's pick $g(x) = x + x^2 + x^3$ and for $f$ let's pick $f(x, y) = x + y$. For $h$ let's pick $h(x) = x - x^2 + x^3$, which agrees with the compositional inverse modulo degree $4$ terms, so $g(h(x) + h(y))$ is a $3$-bud. Expanding, this is (modulo degree $4$ terms)

$$x + y + 2xy + x^2 y + y^2 x.$$

But this shouldn't give a formal group law. (I tried to check this in Sage but it's not happy with expanding the associator here.)