How to think of a Courant algebroid?

Question

I am trying to read read the paper Courant Algebroids and Strongly Homotopy Lie Algebras by Dmitry Roytenberg, Alan Weinstein.

Let $M$ be a smooth manifold. A Courant algebroid on $M$ is a vector bundle $\pi:E\rightarrow M$ with the folllowing extra data:

1. There is a binary operation (skew-symmetric) $[-,-]:\Gamma(M,E)\times \Gamma(M,E)\rightarrow \Gamma(M,E)$ (just like the case of Lie algebroid, but, it need not satisfy the Jacobi identity).
2. There is a bilinear form $\left<-,-\right>:\Gamma(M,E)\times \Gamma(M,E)\rightarrow C^\infty(M)$ non-degenerate, symmmetric) this is somewhat new for me, I know the notion of "metric on vector bundles", but did not use it yet).
3. There is a map $\mathcal{D}:C^\infty(M)\rightarrow \Gamma(M,E)$. It would be slightly less surprising if the map $\mathcal{D}$ is from $C^\infty(M)$ to $\Gamma(M,E^*)$. This is because, there is one obvious way to assign an element of $\Gamma(M,E^*)$ for an element in $C^\infty(M)$, namely $f\mapsto df\mapsto \pi^* (df)$ for $f\in C^\infty(M)$. It is mentioned that, the ''bilinear form'' gives an isomorphism $\Gamma(M,E^*)\rightarrow \Gamma(M,E)$, and the map $\mathcal{D}$ is just the composition $C^\infty(M)\rightarrow \Gamma(M,E^*)\rightarrow \Gamma(M,E)$ (upto a small correction by a constant). This map $\mathcal{D}$ is completely determined by the bilinear form $\left<-,-\right>$. So, this is not actually a separate data. It is coming from the second data mentioned above.

These three data are asked to satisfy some conditions. In total, there are $5$ conditions, out of which $2$ are expected, $1$ is fair enough, the other $2$ are surprising and unmotivated (in first few readings).

1. It is already mentioned that $[-,-]:\Gamma(M,E)\times \Gamma(M,E)\rightarrow \Gamma(M,E)$ does not satisfy the Jacobi condition. It would be useful to know how bad the situation is (how much the Jacobi condition is deviated). The first condition says the situation is not bad, and the Jacobi condition is controlled in a nice way with the help of the map $\mathcal{D}:C^\infty(M)\rightarrow \Gamma(M,E)$. We have $$[[\alpha_1,\alpha_2],\alpha_3]+[[\alpha_2,\alpha_3],\alpha_1]+[[\alpha_3,\alpha_1],\alpha_2]=\mathcal{D}(T(\alpha_1,\alpha_2,\alpha_3)),$$ where $T(\alpha_1,\alpha_2,\alpha_3)\in C^\infty(M)$ is a nice function produced from $[-,-]$ and $\left<-,-\right>$.
2. In case of Lie algebroid, we have said the morphism $\rho^\#:\Gamma(M,A)\rightarrow \Gamma(M,TM)$ is a morphism of Lie bracket. In case of Courant algebroids, we can't ask the map $\rho^\#:\Gamma(M,E)\rightarrow \Gamma(M,TM)$ to be Lie bracket preserving, but, we can ask that the map $\rho^\#:\Gamma(M,E)\rightarrow \Gamma(M,TM)$ preserves the binary operations on $\Gamma(M,E)$ and $\Gamma(M,TM)$, namely $$\rho^\# ([\alpha_1,\alpha_2])=[\rho^\#\alpha_1,\rho^\#\alpha_2],$$ for all $\alpha_1,\alpha_2\in \Gamma(M,E)$.
3. In case of Lie algebroid, we ask that the action of $C^\infty(M)$ on $\Gamma(M,A)$ is compatible with the Lie bracket operation on $\Gamma(M,A)$; that is, we have $$[\alpha, f\alpha_2]=f[\alpha_1,\alpha_2]+(\rho(\alpha_1) f)\alpha_2,$$ for all $\alpha_1,\alpha_2\in \Gamma(M,A)$. In case of Courant algebroid, we ask the same thing, with a slight correction term that comes from the maps $\left<-,-\right>$ and $\mathcal{D}$. We have, $$[\alpha, f\alpha_2]=f[\alpha_1,\alpha_2]+(\rho(\alpha_1) f)\alpha_2 \color{red}{-\left<\alpha_1,\alpha_2\right>\mathcal{D}(f)},$$ for all $\alpha_1,\alpha_2\in \Gamma(M,E)$ and $f\in C^\infty(M)$.
4. some strange condition that says $\left<\mathcal{D}(f),\mathcal{D}(g)\right>=0$ for all $f,g\in C^\infty(M)$.
5. another strange condition that says $$\rho(\alpha)\left<\beta_1,\beta_2\right> =\left<[\alpha,\beta_1]+\mathcal{D}\left<\alpha,\beta_1\right>,\beta_2\right>+ \left<\beta_1,[\alpha,\beta_2]+\mathcal{D}\left<\alpha,\beta_2\right>\right>$$ for all $\alpha,\beta_1,\beta_2\in \Gamma(M,E)$. There is a slight symmetry between first part and second part of the right side. Suppose we take three elements of $\Gamma(M,A)$, two of them combine to give an element in $C^\infty(M)$, one of them gives an element in $\Gamma(M,TM)=\mathfrak{X}(M)$. One can apply the vector field $\rho(\alpha)\in \mathfrak{X}(M)$ on smooth map $\left<\beta_1,\beta_2\right>\in C^\infty(M)$ to get an element $\rho(\alpha)\left<\beta_1,\beta_2\right>\in C^\infty(M)$. One way to get an element in $C^\infty(M)$ from elements of $\Gamma(M,E)$ is by tactically applying $\left<-,-\right>$. I think that is what is done in this case.

I am having difficulty in understanding the conditions 4 and 5. Any extra insight will be helpful. Please let me know any freely available materials that discuss these conditions of Courant algebroids in detail.