EDIT: The answer is trivially positive: see this answer.
Can a non-orientable closed surface of odd genus be immersed in $\mathbb R^3$ so that the associated height function be of Morse-Bott type and have no centers?
That is, the height function would have only Bott-type (circles) extrema and saddle singularities. A Bott-type singularity is a non-degenerate singular circle: a circle where the derivative is zero with the function being quadratic on transverse curves. A center is a Morse-type local extremum: an isolated singularity around which the function is $\pm(x_1^2+x_2^2)$ in some local coordinates.
My intuition is that no. Consider the projective plane $\mathbb RP^2$ as the Boy surface (left) [EDIT: wrong, both figures are not immersions, and the left figure is not the Boy surface] and the Klein bottle $K^2$ (right):
(image from the book). The 8-shaped level sets are immersions (i.e., not self-intersecting) except for where singular points are shown. The vertical line in the right-side figure is a homologically non-trivial cycle.
For even genera $g$ (except $g=2$, which is a different story), it is easy to do: e.g., connect the top and bottom of $K^2$ (right) by a tube (as if you drill a wormhole along the vertical axis), which will form a surface of genus $g=4$ immersed [EDIT: wrong, this is not an immersion] with two Bott-type extrema (circles) and two Morse-type saddles. (You can get any even genus $g\ge4$ by adding more handles.)
However, adding such a handle to the $\mathbb RP^2$ (left) seems not possible. Suppose you add such a handle connecting the bottom to the top of the figure (left). There must be a singularity on that handle. Indeed, consider the evolution of the level sets from the bottom to the top along this handle. The level sets at its endpoints are circles $S^1$ immersed in the plane: O-shaped at the bottom and 8-shaped at the top, which are not regularly homotopic by the Whitney–Graustein theorem. Therefore, there must be a singularity in between.
My intuition is that the singularity will be similar to the saddle shown on the picture: though the singular level set can be more complicated (e.g., connecting more handles), it will effectively convert the left-side picture into the right-side one: it would cause an additional cycle (like the one shown on the right between the two singularities), thus making the genus $g$ even.
I think this argument would generalize to a surface with more handles, as soon as any cycle exists between the "bottom" and "top" of the singular level of the type shown in the figure (left).
Unfortunately, I lack the skill to convert this into a formal proof, and even if I could do it for this particular type of immersion [EDIT: wrong, this is not an immersion] of $\mathbb RP^2$ (Boy surface), it would not prove the claim in the general case. Could you provide such a proof, or point to sources where a proof can be found? Detailed explanations would be greatly appreciated, since I am not an expert.