The Lie algebra of vector fields preserving a tensor

Question

Let $\alpha$ be a tensor field on a manifold $M$, compact and without boundary. Denote by $\mathrm{Diff}(M,\alpha)$ the group of diffeomorphisms of $M$ preserving $\alpha$, i.e. $f:M\to M$ such that $f^*\alpha=\alpha$.

I am interested in the Lie algebra of this group; the tangent space at the identity is given by a subset of vector fields, those satisfying $\mathcal{L}_X\alpha=0$.

I am trying to prove that this set of vector fields is a Lie subalgebra of the vector fields on $M$; in other words, I want to show that if $X$ and $Y$ preserve $\alpha$ then also $\mathcal{L}_{[X,Y]}\alpha=0$. One possible way to do it is to express the flow of $[X,Y]$ using the flows of $X$ and $Y$, as $$\mathcal{L}_{[X,Y]}\alpha=\partial_{t=0}\Phi^{[X,Y]}_t{}^*\alpha.$$ It might be difficult to express $\Phi^{[X,Y]}_t$ as a combination of $\Phi^X_t$ and $\Phi^Y_t$, but in fact we just have to do it to first order in $t$, as is explained in this question, for example. Writing $$\Phi_t^{[X,Y]}=\Phi^X_{\sqrt{t}}\circ\Phi^Y_{\sqrt{t}}\circ\Phi^X_{-\sqrt{t}}\circ\Phi^Y_{-\sqrt{t}}+O(t^2)$$ it seems to follow easily that $\partial_{t=0}\Phi^{[X,Y]}_t{}^*\alpha=0$, since each of the flows preserves $\alpha$. So, a first question is: is this reasoning correct?

I am however interested in proving the identity directly, through a computation in local coordinates. This is motivated by the fact that, in the context of symplectic geometry, it is possible to directly compute in coordinates for example that Hamiltonian vector fields form a Lie algebra, even showing identites like $[X^\omega_f,X^\omega_g]=X^\omega_{\{f,g\}}$ (up to a sign depending on your conventions).

So, let's say that $\alpha$ is a closed $2$-form, and $X$ and $Y$ are vector fields such that $\alpha_{ij}X^i\mathrm{d}x^j$ and $\alpha_{ij}Y^i\mathrm{d}x^j$ are closed; how do I compute that also $$\alpha_{ij}\left(X^k\partial_k Y^i-Y^k\partial_kX^i\right)\mathrm{d}x^j$$ is closed?

As $\alpha_{ij}\partial_kX^i$ is symmetric in $j$ and $k$ (and the same holds for $Y$), we have $$\alpha_{ij}X^k\partial_k Y^i=\partial_j(X^k Y^i)\alpha_{ik}+\alpha_{ij}Y^k\partial_kX^i$$ so that $$\alpha_{ij}\left(X^k\partial_k Y^i-Y^k\partial_kX^i\right)\mathrm{d}x^j=\partial_j(X^k Y^i)\alpha_{ik}\mathrm{d}x^j.$$ At this point however I am unable to check that the differential of this expression vanishes. We get then at my second question: can you proceed from here to show that $\mathrm{d}\left([X,Y]\lrcorner\alpha\right)=0$?