$d_{\nabla}^2u = F_{\nabla}\circ u$? Let $(E,\nabla)\to X$ be a vector bundle endowed with a connection. Let $u\in\Gamma(End(E))$ an endomorphism of $E$. We denote by $d_{\nabla}$ the exterior covariant derivative, induced by $\nabla$ on $\Omega^*(X,E)$ and on $\Omega^*(X,End(E))$. I saw somewhere that
$$d_{\nabla}^2u = F_{\nabla}\circ u$$
but I personally find something like
$$d_{\nabla}^2u=[F_{\nabla},u].$$
What is the right formula?
 A: I guess the formula that you have looked up applies to $\Omega^*(M,E)$ and there you should get $F_{\nabla}\circ s$ in the sense of $(\xi,\eta)\mapsto F_{\nabla}(\xi,\eta)(s)$ for $s\in\Gamma(E)$.
Now applied correctly, this exactly leads to what you have obtained. This is because you have to pass from the connection on $E$ to the induced connection on $End(E)$. Calling this $\tilde\nabla$, the definitions easily imply that for the curvature you get $F_{\tilde\nabla}(\xi,\eta)(u)=[F_{\nabla}(\xi,\eta),u]$. From above you thus conclude that $d_{\tilde\nabla}^2u=F_{\tilde\nabla}\circ u=[F_{\nabla},u]$.
A: In full generality, the formula looks as follows:

Let $E$ be a smooth vector bundle over some smooth manifold
$\mathcal{M}$ and
$\nabla:\mathfrak{X}(\mathcal{M})\times\Gamma(E)\to\Gamma(E)$ be a
connection on $E$ with curvature
$F^{\nabla}\in\Omega^{2}(\mathcal{M},\mathrm{End}(E))$. Then the
exterior-covariant derivative
$\mathrm{d}_{\nabla}:\Omega^{k}(\mathcal{M},E)\to\Omega^{k+1}(\mathcal{M},E)$
satisfies
$$\mathrm{d}_{\nabla}\circ\mathrm{d}_{\nabla}=F^{\nabla}, $$
where $d_{\nabla}\circ d_{\nabla}:\Omega^{k}(\mathcal{M},E)\to
 \Omega^{k+2}(\mathcal{M},E)$ and the right-hand side has to be
understood as  $$F^{\nabla}(\omega):=F^{\nabla}\wedge\omega$$ for all
$\omega\in\Omega^{k}(\mathcal{M},E)$.

The wedge-product which we take here is the wedge product of the form
$$\wedge:\Omega^{2}(\mathcal{M},\mathrm{End}(E))\times\Omega^{k}(\mathcal{M},E)\to\Omega^{k+2}(\mathcal{M},E).$$
These products are defined in the obvious way, i.e.combining the general wedge product and the action of an endomorphism on $E$. More precisely and generally: We define a wedge product of the type
$$\wedge:\Omega^{k}(\mathcal{M},\mathrm{Hom}(E,F))\times\Omega^{l}(\mathcal{M},E)\to\Omega^{k+l}(\mathcal{M},F)$$
for two vector bundles $E,F$ in the following way: For each point $p\in\mathcal{M}$, we define a pairing
\begin{align*}\langle\cdot,\cdot\rangle_{p}:\mathrm{Hom}(E_{p},F_{p})\times E_{p}&\to F_{p}\\ (\varphi,v) &\mapsto \varphi(v).\end{align*}
This gives rise to a well-defined smooth map $\langle\cdot,\cdot\rangle:\mathrm{Hom}(E,F)\times E\to F$ defined point-wise. Then, the wedge product $\wedge$ is the wedge product induced by this pairing, i.e. its is for all $\omega\in\Omega^{k}(\mathcal{M},\mathrm{Hom}(E,F))$ and for all $\alpha\in\Omega^{l}(\mathcal{M},E)$ defined via
\begin{align*}(\omega\wedge\alpha)_{p}(v_{1},\dots,v_{k+l}):=\frac{1}{k!l!}\sum_{\sigma\in\mathfrak{S}^{k+l}}\operatorname{sgn}(\sigma)\langle(\omega_{p}(v_{\sigma(1)},\dots,v_{\sigma(k)}),\alpha_{p}(v_{\sigma(k+1)},\dots,v_{\sigma(k+l)})\rangle_{p}\end{align*}
for all $p\in\mathcal{M}$ and for all $v_{1},\dots,v_{k+l}\in T_{p}\mathcal{M}$. In the case $E=F$, we get the wedge product on $\mathrm{End}(E)$, as defined above.
Now, in your special case, you ask how $\mathrm{d}_{\nabla}\circ\mathrm{d}_{\nabla}$ acts on a zero-form, i.e. on some $u\in\Gamma(E)=\Omega^{0}(\mathcal{M},E)$. In this case, you immediatly see that the wedge-product reduces to
$$F^{\nabla}\wedge u=F^{\nabla}u$$
where the right-hand side $F^{\nabla}u\in\Omega^{2}(\mathcal{M},E)\cong L^{2}_{\mathrm{alt}}(\mathfrak{X}(\mathcal{M}),\Gamma(\mathcal{M},E))$ has to be understood as
$$(F^{\nabla}u)(X,Y):=F^{\nabla}(X,Y)(u)$$
for all vector fields $X,Y\in\mathfrak{X}(\mathcal{M})$.
