Fiber bundle with null-homotopic fiber inclusion

Here is how to construct a map from $\pi_{n-1}(F)\to \pi_n(B)$. Given a sphere $f\colon S^{n-1}\to F$, by hypothesis it bounds a disk $g\colon D^{n}\to E$. Consider the projection $\pi\circ g\colon D^n\to B$, where $\pi\colon E\to B$ is the projection map from the fiber bundle. Note that $\pi\circ g(\partial D^n)=\{*\}$ is a single point, so it represents a map of $S^{n}$ into $B$. I.e. it gives an element of $\pi_{n}(B)$ as desired. Now you have to show this map gives a well-defined homomorphism and is a splitting.

Edit: Responding to OP's comments, the right-hand splitting is the one you need. I doubt there is a natural left-hand splitting. The way you get the boundary operator in the long exact sequence is to take a map $f\colon S^n\to B$ representing your element of $\pi_n(B)$. You can think of it as a map $(D^n,\partial D^n)\to B$, where $\partial D^n$ maps to a point. Now by the homotopy lifting property for fiber bundles, you can lift $f$ to a map $\tilde{f}\colon D^n\to E$, but now $\tilde{f}(\partial D^n)$ will not be a point, but will instead lie in $\pi^{-1}(*)=F$. So $\tilde{f}$ give you an element of $\pi_{n-1}(F)$ as desired.