Properties of Lie derivative Let's have Lie derivative:
$$
L_{V}\varphi = V^{\mu}\partial_{\nu}\varphi , \quad L_{V}A_{\mu} = V^{\nu}\partial_{\nu}A_{\mu} + (\partial_{\mu}V^{\nu})A_{\nu}. 
$$
How to show that for scalar and vector fields
$$
L_{V}L_{U} - L_{U}L_{V} = L_{[U, V]}, [U, V]^{\nu} = U^{\mu}\partial_{\mu}V^{\nu} - V^{\mu}\partial_{\mu}U^{\nu}?
$$
An attempt to prove equality for vector field.
$$
L_{[U, V]}A_{\mu} = [U, V]^{\nu}\partial_{\nu}A_{\mu} + (\partial_{\mu}[U, V]^{\nu})A_{\nu} = 
$$
$$
=[U, V]^{\nu}\partial_{\nu}A_{\mu} + (\partial_{\mu}(U^{\alpha}\partial_{\alpha}V^{\nu} - V^{\alpha}\partial_{\alpha}U^{\nu}))A_{\nu}. \qquad (.1)
$$
By the other way [edited],
$$
(L_{U}L_{V} - L_{V}L_{U})A_{\mu} = L_{U}(V^{\nu}\partial_{\nu}A_{\mu} + \partial_{\mu}(V^{\nu})A_{\nu}) - L_{V}(U^{\nu}\partial_{\nu}A_{\mu} + \partial_{\mu}(U^{\nu})A_{\nu})=
$$
$$
= U^{\alpha}\partial_{\alpha}[V^{\nu}\partial_{\nu}A_{\mu} + \partial_{\mu}(V^{\nu})A_{\nu}] + \partial_{\mu}(U^{\alpha})[V^{\nu}\partial_{\nu}A_{\alpha} + \partial_{\alpha}(V^{\nu})A_{\nu}] -
$$
$$
-V^{\alpha}\partial_{\alpha}[U^{\nu}\partial_{\nu}A_{\mu} + \partial_{\mu}(U^{\nu})A_{\nu}] - \partial_{\mu}(V^{\alpha})[U^{\nu}\partial_{\nu}A_{\alpha} + \partial_{\alpha}(U^{\nu})A_{\nu}]=
$$
$$
=[U, V]^{\nu}\partial_{\nu}A_{\mu} + \partial_{\mu}(U^{\alpha})V^{\nu}\partial_{\nu}A_{\alpha} + U^{\alpha}\partial_{\alpha}(\partial_{\mu}(V^{\nu})A_{\nu}) + \partial_{\mu}(U^{\alpha})\partial_{\alpha} (V^{\nu})A_{\nu} -
$$
$$
- \partial_{\mu}(V^{\alpha})U^{\nu}\partial_{\nu}A_{\alpha} - V^{\alpha}\partial_{\alpha}(\partial_{\mu}(U^{\nu})A_{\nu}) - \partial_{\mu}(V^{\alpha})\partial_{\alpha}(U^{\nu})A_{\nu}. \qquad (.2)
$$
But I don't know how to make $(.2)$ equal to $(.1)$.
Edit.
By rewriting the second summand in $(.1)$ as
$$
\partial_{\mu}(U^{\nu})\partial_{\nu}(V^{\alpha})A_{\alpha} + \partial_{\mu}(U^{\nu})V^{\alpha}\partial_{\nu}A_{\alpha} + U^{\nu}\partial_{\nu}(\partial_{\mu}(V^{\alpha})A_{\alpha}) - \partial_{\mu}(V^{\nu})\partial_{\nu}(U^{\alpha})A_{\alpha} - \partial_{\mu}(V^{\nu})U^{\alpha}\partial_{\nu}A_{\alpha} -V^{\nu}\partial_{\nu}(\partial_{\mu}(U^{\alpha})A_{\alpha})
$$
I got
$$
[L_{[U, V]} - (L_{U}L_{V} - L_{V}L_{U})]A_{\mu} =
$$
$$
=\partial_{\mu}(U^{\nu})V^{\alpha}\partial_{\nu}A_{\alpha} - \partial_{\mu}(V^{\nu})U^{\alpha}\partial_{\nu}A_{\alpha} - \partial_{\mu}(U^{\alpha})V^{\nu}\partial_{\nu}A_{\alpha} + \partial_{\mu}(V^{\alpha})U^{\nu}\partial_{\nu}A_{\alpha},
$$
which isn't equal to zero.
 A: *

*On scalars


$$L_V(L_U(\phi)):=v^{\rho}\partial_\rho u^{\nu}\partial_\nu\phi+
v^{\rho}u^{\nu}\partial^2_{\rho\nu}\phi; $$
$$L_U(L_V(\phi)):=u^{\rho}\partial_\rho v^{\nu}\partial_\nu\phi+
u^{\rho}v^{\nu}\partial^2_{\rho\nu}\phi=(\text{repeated indices})=\\ 
u^{\nu}\partial_\nu v^{\rho}\partial_\rho\phi+
u^{\nu}v^{\rho}\partial^2_{\rho\nu}\phi.$$
$$L_{[U,V]}(\phi):=[U,V]^{\rho}\partial_\rho \phi=
u^{\nu}\partial_\nu v^{\rho}\partial_\rho\phi
-v^{\nu}\partial_\nu u^{\rho}\partial_\rho\phi=(\text{repeated indices})=\\ 
u^{\nu}\partial_\nu v^{\rho}\partial_\rho\phi
-v^{\rho}\partial_\rho u^{\nu}\partial_\nu\phi$$
i.e. $L_U\circ L_V-L_V\circ L_U=L_{[U,V]}.$


*

*On vector fields


$$L_{[U,V]}(A_\mu)=u^{\rho}\partial_\rho v^{\nu}\partial_\nu A_\mu-
v^{\rho}\partial_\rho u^{\nu}\partial_\nu A_\mu+
\partial_\mu(u^{\rho}\partial_\rho v^{\nu}-v^{\rho}\partial_\rho u^{\nu})A_\nu,$$
$$L_U(L_V(A_\mu)):=u^{\rho}\partial_\rho B_{\mu}+
\partial_\mu u^\rho B_\rho,$$
with $B_\mu=v^\nu\partial_\nu A_\mu+\partial_\mu v^\nu A_\nu$ and similarly for $B_\rho$. Similarly
$$L_V(L_U(A_\mu)):=v^{\rho}\partial_\rho C_{\mu}+
\partial_\mu v^\rho C_\rho,$$
with $C_\mu=u^\nu\partial_\nu A_\mu+\partial_\mu u^\nu A_\nu$ and similarly for $C_\rho$.
The identity on vector fields is then proven by a bit long but straightforward computation.


*

*Edit: on the OP computations


Two little remarks. In the computations leading to .2) the last term in the r.h.s of the second equality has a wrong sign (it should be a minus).
The first term in the last line of the same sequence of computations, i.e.  $-V^{\alpha}\partial_{\alpha}(U^{\nu}\partial_{\nu}A_{\mu}) $ should not be.
Removing such term, we arrive at
$$L_{[U,V]}(A_\mu)-L_U(L_V(A_\mu))-L_V(L_U(A_\mu))=
U^{\alpha}\partial_{\mu}V^{\nu}\partial_\alpha A_{\nu}  - V^{\alpha}\partial_{\mu} U^{\nu}\partial_{\alpha}A_{\nu}+
\\
\partial_{\mu}U^{\alpha}V^{\nu}\partial_\nu A_{\alpha}
-\partial_{\mu}V^{\alpha} U^{\nu}\partial_{\nu}A_{\alpha};$$
the r.h.s. of the above equality is equal to $0$: one needs to exchange the repeated indices $\nu\leftrightarrow \alpha$ in the last 2 terms to finish the proof.
