Suppose that $\pi:E \to D$ is a 4-dimensional Lefschetz fibration over a disk, and let $\Omega$ be a closed 2-form on $E$ such that it is non-degenerate fiberwise. 


For any $x \in E$, there is a decomposition $TE\_x=TE\_x^h \oplus TE\_x^v$, $TE^v=\ker d\pi$ and $TE\_x^h=\{v \in TE\_x: \Omega(v,w)=0,\forall w \in TE\_x^v\}$. Let $\partial\_t$ be the coordinate vector field on $S^1=\partial D$, $R$ be the horizontal lift of $\partial\_t$ and $\phi\_{\Omega}$ be the flow generated by $R$. Taking base point $1 \in S^1$, then we get a map $\phi\_{\Omega}: \pi^{-1}(1)\to\pi^{-1}(1)$, so the boundary $\partial E=\pi^{-1}(\partial D)$ is a mapping torus $Y\_{\phi\_{\Omega}}$.


For the mapping torus, we can define a periodic orbit to be an integral curve of vector field $R$. I don't know whether the periodic orbit are non degenerate. 



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> Can I do a perturbation $\Omega'$ of $\Omega$ such that the periodic orbits with period $\le d$ on the mapping torus $Y\_{\phi\_{\Omega'}}=\pi^{-1}(\partial D)$ are non-degenerate?
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Non-degenerate periodic orbits in the boundary of Lefschetz fibration over a disk
This is very rough and probably not too precise but here is what I
think.


For a two-form $\omega$ on a manifold $M$ denote by
$\text{null}(\omega\_x)$ the vector subspace of the tangent space
$T\_xM$ defined as follows:
$$\text{null}(\omega\_x) = \left\{ \, W \in T\_xM \,\, | \,\, \omega\_x(W,v) = 0
\, \text{ for all } \, v \in T\_xM \, \right\},$$ i.e. the
null-space of $\omega$ contains all tangent vectors that are
orthogonal to the whole tangent space $T\_xM$. In the case when
$M=E$ and the assumption that $\Omega$ is non-degenerate
fiber-wise, $\dim{\text{null}(\Omega)} = 2$ or $0$. When the
dimension of the null-space is $0$ then $\Omega$ is non-degenerate
on $E$ and hence it is symplectic. Otherwise, it is symplectic on
the fibers only. However, when restricted to the even-dimensional
(more precisely three-dimensional) fibration over the unit circle,
the null-space is one-dimensional in both cases, so it defines a
line field.


Let $z = y\_1 + i y\_2$ be a notation for the points on the disc $D
= \{ z \in \mathbb{C} \, : \, |z| < 1 + \varepsilon \}$. In polar
coordinates we have $z = R e^{i\theta}$ where $R = |z|$ and
$\theta = \text{Arg}(z) \in [0,2\pi)$. Define the functions
$R^2(x) = |\pi(x)|^2$ and $\theta(x) = \text{Arg}(\pi(x))$. For $z
\in D$ let $E\_z = \pi^{-1}(z)$ be the fiber over $z$. Now, the
lift of the standard symplectic form $2 \, dy\_1 \wedge dy\_2$ can be
written as
$$\pi^\*(2 \, dy\_1 \wedge dy\_2) = d (R^2) \wedge d \theta.$$ Denote by $M\_1 = \{ x \in E \, : \, |\pi(x)|^2 = 1 \}$. Take
$$l(x) = \{ W \in T\_xM\_1 \, : \, \Omega\_x(W,v) = 0 \, \text{ for
all } \, v \in T\_x E\_{\pi(x)} \, \}.$$ This is a line field and
is equal to $T\_xE^h \cap T\_xM\_1 = l(x)$, i.e. $l(x)$ is a one
dimensional line field tangent to $M\_1$ and $\Omega$-orthogonal to
the fibers. Observe that you can also write it as the null-space
of $\omega\_x = \Omega\_x - d R^2 \wedge d \theta$. The lift of the
vector field tangent to the unit circle in the $D$ lies on $l(x)$.
Take any function $H : E \to \mathbb{R}$ and restrict it to $M\_1$.
If you look at the form $\Omega'\_x = \Omega\_x - d(R^2) \wedge
\theta - d H \wedge d \theta$ restricted to $M\_1$, then take
its one dimensional null-space $l\_H(x)$. It gives
you a line field on $M\_1$. The vector field on the unit circle
lifts to a vector field whose span defines $L\_H(x)$. Furthermore,
its tangent trajectories give rise to a large family of
symplecotmorphisms (parametrized by $H$) on the fiber
$\pi^{-1}(1)$ by using the constriction of first return-map. Then,
using the null-space of $\omega$, you can trivialize $M\_1$ by
creating the cover $p : E\_1 \times \mathbb{R} \to M\_1$ and as a
result you obtain more or less something like a time-dependent
(actually, $\theta$-dependent) Hamiltonian $\tilde{H}(y,\theta)$
where $(y,\theta) \in E\_1 \times \mathbb{R}$ that is
$2\pi-$periodic and gives rise to a symplectomorphism on $E\_1$ (a
version of the Poincare first return map). Then it looks like
generically, it is possible that its periodic points are isolated.


So a deformation of $\Omega$ that gives rise to non-degenerate
periodic orbits can be obtained by adding $d\theta \wedge \big(R^2
+ H \big)$. Observe that although $\theta$ is not well-defined on
the unit circle, its differential $d \theta$ is, because $d \theta
= d(\theta + 2\pi)$. Furthermore, if you feel uncomfortable about
$d \theta$ around the center of $D$, then multiply your whole
perturbation $d\theta \wedge \big(R^2 + H \big)$ by an infinitely
smooth bump-function which is constantly $1$ everywhere in $D$
except for a very small disc neighborhood around its center and it
is zero at the center.


