I also only learned about Frolicher-Nijenhuis brackets from Saunders' book on jets but I doubt that there is any other authorative reference besides Michor's book and the original papers. I don't know if this is what the original poster intended, but here's how I understand the link between FN and curvature. Generally, I tend to stay away from the full generality of the FN bracket, and only use those (axiomatic) properties that I need...

Let $\pi: E \rightarrow M$ be a fiber bundle. An Ehresmann connection is a subbundle $H$ of $TE$ such that $H \oplus VE = TE$, where $VE$ is the vertical bundle. Denote the projector from $TE$ onto $H$ by $h$. Saunders (and many other authors) define Ehresmann connections directly in terms of bundle maps $h$, since they are easier to work with, but this doesn't matter.

Now, to introduce curvature, we would like to formalize the idea that curvature is the failure of "parallel transport" to commute, where of course we haven't define parallel transport properly. However, it is only a small step of the imagination to guess that this must be related to the integrability of $H$, i.e.whether $[h(X), h(Y)] \in H$ for arbitrary vector fields $X, Y$. The failure of two horizontal vector fields to be horizontal again is measured by the expression
$$
R(X,Y) = [h(X), h(Y)] - h([h(X), h(Y)]),
$$
and in fact this is nothing but Saunders' definition 3.5.13 of curvature. Note that $R(X, Y) = 0$ if either $X$ or $Y$ is in $VE$.

The Frohlicher-Nijenhuis bracket of two linear bundle maps is in general quite complicated, but if you look at proposition 3.4.15 in Saunders, you get that for a linear bundle map $h: TE \rightarrow TE$,
$$
[h, h] (X,Y) = 2(h([X,Y]) + [h(X), h(Y)] - h([h(X),Y]) - h([X,h(Y)])).
$$
Now note that the right-hand side is zero as soon as either $X$ or $Y$ is in $VE$, just as for $R$. If both $X$ and $Y$ are horizontal, we can apply the above formula to $X = h(X)$ and $Y = h(Y)$, and conclude that $[h, h] (X, Y) = 2( [h(X), h(Y)] - h([h(X), h(Y)]))$ (where we use the identity $h \circ h = h$ liberally), and so
$$
R = \frac{1}{2} [h,h].
$$

If you have a principal fiber bundle, $h$ is related to the connection one-form $\mathcal{A}$, and the above formula gives the curvature of $\mathcal{A}$ in terms of the covariant differential of $\mathcal{A}$. For a symplectic connection, something similar happens.

**Edit:** here's how I think it works for principal fiber bundles. Take a connection one-form $A: TE \to \mathfrak{g}$, and let $\sigma: \mathfrak{g} \rightarrow VE$ be the infinitesimal generator of the $G$-action. The composition $v := \sigma \circ \mathcal{A}$ is then the vertical projector of the connection and $h := 1 - v$ is the horizontal one. Now plug this expression for $h$ into the formula for the curvature:
$$
[h, h] = [1, 1] - [\sigma \circ \mathcal{A}, 1] - [1, \sigma \circ \mathcal{A}]
+ [\sigma \circ \mathcal{A}, \sigma \circ \mathcal{A}].
$$
The term $[1,1]$ is zero, and you can use Saunders' proposition 3.4.15 to show that term 2 and 3 vanish as. The last term can be written as (again using S3.4.15) as
$$
[\sigma \circ \mathcal{A}, \sigma \circ \mathcal{A}] (X, Y) =
2( (\sigma \circ \mathcal{A})^2([X, Y]) + [ (\sigma \circ \mathcal{A})(X), (\sigma \circ \mathcal{A})(Y)] - (\sigma \circ \mathcal{A})([(\sigma \circ \mathcal{A})(X), Y)]) - (\sigma \circ \mathcal{A})([X, (\sigma \circ \mathcal{A})(Y)])).
$$

Now, show that this vanishes whenever $X$ or $Y$ is vertical, so that we can take $X$ and $Y$ to be horizontal. In that case, the above simplifies to
$$
[\sigma \circ \mathcal{A}, \sigma \circ \mathcal{A}] (X, Y) = 2(\sigma \circ \mathcal{A})([X, Y])
$$
or
$$
R(X, Y) = (\sigma \circ \mathcal{A}) ([X^h, X^h])
$$
where $X^h$ represents the horizontal part of $X$. But $\mathcal{A} ([X^h, X^h])$ is just the negative of the curvature (as a two-form with values in $\mathfrak{g}$) of $\mathcal{A}$, so that
$$
R(X, Y) = -\sigma ( \mathcal{B}(X, Y) ).
$$

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