Frölicher–Nijenhuis bracket

Remarks

The bracket exists and is unique because every graded derivation of $\Omega(M)$ that commutes with $d$ in the graded sense is $\mathcal{L}_K$ for exactly one vector-valued form $K$. It is graded antisymmetric, $[K, L] = -(-1)^{kl} [L, K]$, and satisfies the graded Jacobi identity.

For $\alpha \in \Omega^k(M)$, $\beta \in \Omega^l(M)$ and vector fields $X, Y$, an explicit formula is $$[\alpha \otimes X, \beta \otimes Y] = \alpha \wedge \beta \otimes [X, Y] + \alpha \wedge \mathcal{L}_X \beta \otimes Y - \mathcal{L}_Y \alpha \wedge \beta \otimes X + (-1)^k \big( d\alpha \wedge \iota_X \beta \otimes Y + \iota_Y \alpha \wedge d\beta \otimes X \big).$$

Example: for $k = l = 0$, take $\alpha = \beta = 1$. The formula gives $[X, Y]$, the Lie bracket of vector fields, and $\mathcal{L}_X$ is the usual Lie derivative. Another example: the identity $\mathrm{Id} \in \Omega^1(M, TM)$ has $\iota_{\mathrm{Id}} \omega = j \omega$ for $\omega \in \Omega^j(M)$. Hence $\mathcal{L}_{\mathrm{Id}} \omega = (j + 1) d\omega - j\, d\omega = d\omega$, so $\mathcal{L}_{\mathrm{Id}} = d$ and $[\mathrm{Id}, L] = 0$ for every $L$.

Non-example: the naive rule $[\alpha \otimes X, \beta \otimes Y] = \alpha \wedge \beta \otimes [X, Y]$ does not define a bracket on $\Omega(M, TM)$. For a function $f$, the elements $f\alpha \otimes X$ and $\alpha \otimes fX$ are equal, but the rule gives different results for them, since $[fX, Y] \ne f[X, Y]$ in general.