Let $\mathfrak{g}$ be a finite dimensional simple Lie algebra over $\mathbb{C}$.

The polynomial current Lie algebra $\mathfrak{g}[t] = \mathfrak{g} \otimes \mathbb{C} [t]$
has the bracket
$$[xt^r, yt^s] = [x,y] t^{r+s}$$
for $x,y \in \mathfrak{g}$. It is graded with deg$(t) = 1$.

If we set $h=0$ in Drinfeld's first presentation of the Yangian (given in Theorem 12.1.1 of Chari and Pressley's Guide to Quantum Groups) then we get a presentation of $U(\mathfrak{g}[t])$ where the generators are the elements $x \in \mathfrak{g}$ and $J(x) = xt$ of $\mathfrak{g}[t]$ with degree $=0,1$, and the relations all have degree of both sides less than $3$.

Specifically we require that all the relation in $\mathfrak{g}$ are satisfied for the elements with degree 0, and (for all $x,y, x_i, y_i, z_i \in \mathfrak{g}$ and complex numbers $\lambda, \mu$):

$$\lambda xt + \mu yt = (\lambda x + \mu y)t$$ $$[x, yt] = [x,y]t,$$ $$\sum_i [x_i, y_i] = 0 \implies \sum_i [x_i t, y_i t ] = 0$$ $$ \sum_i [[x_i, y_i], z_i] = 0 \implies \sum_i [[x_i t, y_i t], z_i t]=0$$ Then assuming that all the relations of degree less than or equal to $3$ hold is enough to get the remaining ones. The elements $xt^2, xt^3, \ldots$ are defined inductively. This can be proved by induction, using the Serre presentation of the finite-dimensional Lie algebra and then checking all the required relations in several cases. But even in the $\mathfrak{sl}_2$ case the argument is laborious.

Is there a better way of seeing that one needs only relations of degree less than three in order to get the rest?