Let $r$ and $k$ denote, respectively, a polynomial degree and the degree of differential forms over $\mathbb R^n$. For $r,k \ge 0$ with $r+k > 0$, denoting by $\kappa$ the Koszul differential, by $\rm d$ the usual differential of the De Rham complex, and by $\mathcal P_r\Lambda^k$ the space of polynomial differential $k$-forms of degree $r$ over $\mathbb R^n$, the following decomposition holds (cf. Arnold et al., Acta Numerica, 2006, equation (3.11)):
$$\mathcal P_r\Lambda^k = \kappa \mathcal P_{r-1}\Lambda^{k+1} \oplus \mathrm{d} \mathcal P_{r+1}\Lambda^{k-1}.$$
Now, let $k=0$ and $r \ge 1$, so that assumption $r+k > 0$ is verified. The second term in the above direct sum is then $\{0\}$, as $k-1 = -1 < 0$. This means that
$$ \mathcal P_r\Lambda^0 = \kappa \mathcal P_{r-1}\Lambda^{1},\quad\forall r \ge 1.$$
I have some troubles figuring out how can this be true. Reasoning in terms of proxies, $1$-forms can be identified with vector fields, so that applying $\kappa$ to this space means taking the scalar product of a vector of polynomials of degree $r$ with the identity vector field over $\mathbb R^n$. However, I am not sure this gives the entire space of scalar polynomials of degree $r$ over $\mathbb R^n$, since taking the scalar product with $x\in\mathbb R^n$ rules out constants.
For instance, for $n=2$, $r=1$ and the vector of constants $(a,b)\in \mathcal P_0\Lambda^1(\mathbb R^2) \simeq \mathbb R^2$, taking the scalar product with $(x_1,x_2)\in\mathbb R^2$ yields $ax_1+bx_2$. To fill the entire space of scalar polynomials of degree $1$ over $\mathbb R^2$, one would need to add up a constant to the previous expression.
Am I missing something? Is there any link with the fact that the origin in $\mathbb R^n$ is arbitrary in order to define the Koszul differential?
You are not missing anything, Francesco. If you restrict to the space of homogeneous polynomial $k$-forms of degree $q$, $\mathcal H_q\Lambda^k$, then the decomposition $$\mathcal H_q\Lambda^k = \kappa \mathcal H_{q-1}\Lambda^{k+1} \oplus \mathrm{d} \mathcal H_{q+1}\Lambda^{k-1} \tag{1}\label{1}$$ holds for all $k,q\ge0$, except for $k=q=0$, in which case the left-hand side is $\mathbb R$ but the right-hand side is zero.
Now let $r\ge 1$ and sum \eqref{1} up over $0\le q\le r$. This gives the equation you quoted, $$ \mathcal P_r\Lambda^k = \kappa \mathcal P_{r-1}\Lambda^{k+1} \oplus \mathrm{d} \mathcal P_{r+1}\Lambda^{k-1}, \tag{2}\label{2} $$ so long as $k>0$. But if $k=0$, there is the discrepancy you noticed, since \eqref{1} is not true for $k=q=0$. In this case, the right-hand side of \eqref{2} is equal to the space of polynomials of degree at most $r$ for which the constant term vanishes.