Consider the restriction $f$ of the map $F:\mathbb R^3\rightarrow \mathbb R,(x,y,z)\mapsto z$ to $S^2$ :
indeed it is a smooth function and according the method of Lagrange multiplier, you know that $f$ has only two critical points (no choice, it has to be the max and min of $f$) located at $p_\epsilon=(0,0,\epsilon)$, $\epsilon=\pm 1$.
To find the nature of this two points, you need to find a chart around $p_\epsilon$ and compute the Hessian of $f$.
Around $p_\epsilon$, the map $\varphi : (x,y)\in B(0,r) \mapsto (x,y,\epsilon\sqrt{1-x^2-y^2})$ is nice chart and $f\circ \varphi (x,y)=\epsilon\sqrt{1-x^2-y^2}$ on a neighbourhood of $p$. The study of the map $g=f\circ \varphi$ around $(0,0)$ will give you the nature of $p_\epsilon$.
You can check that $$\dfrac{\partial^2 g}{\partial x^2}(0,0)=\dfrac{\partial^2 g}{\partial y^2}(0,0)=-\epsilon$$ and $$\dfrac{\partial^2 g}{\partial x \partial y}(0,0)=0.$$
It means that the Hessian of $f$ at $p_\epsilon$ is either negative definite or positive definite according to $\epsilon =1$ or $\epsilon =-1$. In other words $f$ looks like $(x,y)\mapsto f(p_\epsilon)-\epsilon(x^2+y^2)$ around $p_\epsilon$ where $(x,y)$ are local coordinates and the number of negative square is the Morse index (according to Morse lemma).
So for $\epsilon =1$, $p_\epsilon=(0,0,1)$ is a maximum and its index is equal to $2$.
For $\epsilon = -1$, $p_\epsilon=(0,0,-1)$ is a minimum and its index is equal to $0$.
Finally, if I denote by $C_i$ the $\mathbb Z/2\mathbb Z$-vector space generated by critical points of $f$ of index $i$, the Morse theory tells you that we have an exact sequence $$\{0\}\rightarrow C_2 \rightarrow C_1 \rightarrow C_0 \rightarrow \{0\}.$$
In this case, $C_0\simeq \mathbb Z/2\mathbb Z \simeq C_2$ and $C_1=\{0\}$, so the Morse cohomology of this complex is $$HM_i=\left\{\begin{array}{ll} \mathbb{Z} / 2 \mathbb Z & \text{if } i=0,i=2 \\ \{0\} & \text{otherwise}\end{array}\right.$$
and $\chi(S^2)=1-0+1=2$.
To show that a compact manifold $V$ of dim $d$ has a lot of Morse functions, one can use the map $f_p(x)=\|x-p\|^2$ (the square gives regularity). Let's think of $V$ as a submanifold of $\mathbb R^n$.
- Check that a point $c\in V$ is a critical point of $f_p$ iff $c-p\perp T_cV$.
- If $v=v(x_1,\cdots,x_d)\in V$ is a local chart around $c$, then $$\dfrac{\partial^2 f}{\partial x_i\partial x_j}=2(<\dfrac{\partial v}{\partial x_i},\dfrac{\partial v}{\partial x_j}>+<v-p,\dfrac{\partial^2 v}{\partial x_i\partial x_j}>).$$ You want to show that for almost every $p$, this matrix is definite i.e. of rank $d$.
- Consider the normal bundle $N=\{(v,w)\in V\times \mathbb R^n | w\perp T_xV\}\subset V\times \mathbb R^n$ and $F:N\rightarrow \mathbb R^n, (v,w)\mapsto v+w$. $N$ is a manifold of dimension $n$ and $F$ is smooth.
- Check that $p=v+w\in \mathbb R^n$ is a regular value of $F$ iff $\forall v,w$ s.t. $p=v+w$ the matrix $M=(m_{ij})$ given by $$m_{ij}=<\dfrac{\partial v}{\partial x_i},\dfrac{\partial v}{\partial x_j}>+<w,\dfrac{\partial^2 v}{\partial x_i\partial x_j}>$$ is invertible.
With this four points checked, apply the Sard lemma and you get the proof.
Transversality helps you to construct in generic way the morse complex. Roughly speaking, to define the complex, you need a nice morse function, a nice pseudo-gradient vector field, etc... that all fit together. It is not obvious that one can always do that but the transversality conditions allows you to say : if it doesn't work at some point, then you can perturb a little bit all this tools so that it will.
Permit me to talk about the Morse complex over $\mathbb {Z}$. For $\mathbb{Z}_2$ a similar argument works.
A function whose Morse complex has trivial boundary maps is called a perfect Morse function. Finding perfect Morse functions is hard; To my best knowledge it is as open problem to characterize manifolds that admit perfect Morse functions.
Simple examples of manifold which will never have perfect Morse functions can be made as follows: Take a group $G$, whose abelianization is trivial. Construct a manifold $M$, with fundamental group $G$ (This is always possible if $G$ is finitely presented group for a manifold $M$ of dimension $4$). Then $H_1(M;\mathbb{Z})=G_{\mathrm{ab}}=0$. A perfect Morse function would not have critical points of index $1$. But since $\pi_1(M)=G$, this cannot be the case.
Best Answer
Take the function $f(x_1, \dots, x_{d}) = x_{d}$ defined on $\mathbb S^{d-1}$. Let $\phi : \mathbb S^{d-1 }\to \mathbb S^{d-1}$ be a diffeomorphism such that $\phi (0,\cdots, 0,1) = (0,\cdots, 0,1)$ and $y:=\phi^{-1}(0,\cdots, 0,-1) \neq (0,\cdots, 0,-1)$. Then the new function $g = f \circ \phi $ has two critical points $(0,\cdots, 0, 1)$ and $y$, and these two critical points are not antipodal.
For a slightly more concrete example: let $d=2$ and let $p = (0,\sqrt 2)$. Let $L^\pm$ be two lines in $\mathbb R^2$ touching $\mathbb S^1$ at $(\pm \sqrt 2^{-1}, \sqrt 2^{-1})$ respectively. rotating the line from $L^-$ to $L^+$ gives you a foliation of $\mathbb S^1$. One can find a function $f: \mathbb S^1 \to \mathbb R$ so that the level set corresponds to this foliation. $f$ would have exactly two critical points $(\pm \sqrt 2^{-1}, \sqrt 2^{-1})$, which is not antipodal.