Let $X,Y$ be smooth manifolds, $S'$ a submanifold of $Y$, and $f:\mathbb{R}\times X\to Y$ a smooth function. Generically, we have that $f$ is transverse to $S'$, which implies that $S:=f^{-1}(S')$ is a smooth submanifold of $\mathbb{R}\times X$. It comes with a projection map, namely the restriction of the projection on $\mathbb{R}$ to $S$. My question is:
Is there a generic property I can also ask $f$ to check in order for this projection to be a Morse function?
So far analyzing this first transversality condition, I can describe the tangent space of $S$, $$T_{(t,x)}S=\left\{(\dot{t},\dot{x})\in \mathbb{R}\times T_xX\,;\,(d_Xf)_{(t,x)}(\dot{x})+\dot{t}\frac{\partial f}{\partial t}(t,x)\in T_{f(t,x)}S'\right\},$$ and thus if $(t,x)$ is a critical point of $\pi|_S$, then every tangent vector of $S$ at $(t,x)$ is sent to $0$ by $d\pi_{(t,x)}$, which amounts to say that $$T_{(t,x)}S=\left\{(0,\dot{x})\in \mathbb{R}\times T_xX\,;\,(d_Xf)_{(t,x)}(\dot{x})\in T_{f(t,x)}S'\right\}\simeq(d_Xf)_{(t,x)}^{-1}(T_{f(t,x)}S').$$ From this we can say that $f_t$ is not transverse to $S'$, otherwise we would have a problem of dimension ($T_{(t,x)}S$ would be 1 dimension less than expected). So by the transversality parametric theorem, I can conclude that the regular values of the projection are generic. But there is nothing about the critical values being non-degenerate, and I can even construct a counter-example where the transversality condition alone does not result in the projection being Morse: take $f:\mathbb{R}\times\mathbb{R}\to \mathbb{R}^2$ given by $f(t,x)=(t,x)$ and $S'=g(\mathbb{R})$ with $g(x)=(x^3,x)$. So I have to add something else, probably a transversality condition involving the 1-jet of $f$, but I can't find it and every help will be much appreciated. Thanks for reading my question!
Edit regarding Ted Shifrin's comment: Taking a metric on $S$, the fact that $\pi$ is Morse translates as $\mathrm{grad}\,\pi$ is transverse to $0_{TS}$, or again that for all $(t,x)\in S$ s.t. $d\pi_{(t,x)}=0$, then $$\nabla\mathrm{grad}\,\pi:T_{(t,x)}S\to T_{(t,x)}S,(0,\dot{x})=\dot{x}\mapsto\nabla_{\dot{x}}\mathrm{grad}\,\pi$$ is onto. Since in coordinates, $\nabla_{\dot{x}}\mathrm{grad}\,\pi$ is given by $$\left(\dot{x}^i\partial_l\pi(\partial_ig^{lk}+g^{lj}\Gamma_{lj}^k)\right)\frac{\partial}{\partial x^k},$$ this amounts to show that for all $\dot{y}=\dot{y}^k\frac{\partial}{\partial x^k}$ s.t. $(d_Xf)_{(t,x)}(\dot{y})\in T_{f(t,x)}S'$, there exists a $\dot{x}=\dot{x}^i\frac{\partial}{\partial x^i}$ s.t. $(d_Xf)_{(t,x)}(\dot{x})\in T_{f(t,x)}S'$ and $$\dot{x}^i\partial_l\pi(\partial_ig^{lk}+g^{lj}\Gamma_{lj}^k)=\dot{y}^k.$$ What bothers me from this writing is that I still can't see what property I could add to $f$ in order to this be verified (it will be as soon as $\det\left(\partial_l\pi(\partial_ig^{lk}+g^{lj}\Gamma_{lj}^k)\right)_{ik}\neq 0$, but I think that this is equivalent to ask that $\mathrm{Hess}(\pi)_{(t,x)}$ is non-degenerate, which is simply the definition of $\pi$ being a Morse function ; and I don't manage to involve the condition $(d_Xf)_{(t,x)}(\dot{y})\in T_{f(t,x)}S'$). Again, every help from here will be much appreciated!
