Prove no existing a smooth function satisfying ... related to Morse Theory

Here a counterexample to 2) (without "properness" assumptions)

Let $B$ the ball of center $0$ and radius $1$.

Let $g:B\to\mathbb R$ be a function with property 2) (positive with two minima where the function is zero and no other critical point). This is easily constructed.

Now consider any diffeomorphism $\phi:\mathbb R^2\to B$.

Now the function $f=g(\phi(x))$ has the requested properties.

To build $g$ consider a smooth positive function $h$ with isolated critical poitns and two minima $p_1$ and $p_2$ where $h=0$. Now join $p_1$ to $p_2$ with a simple arc which avoids critical points other than the $p_i$. A regular neighborood of such arc is a disc $D$ where $h$ has no critical points other than at $p_1$ and $p_2$. Now uses a diffeomorphism from $B$ to $D$ and compose with $h$.

EDIT from the comment of "This is much healthier": Begin with $g(x,y)=(x^2−1)^2+y^2$, take any simply-connected domain that contains the minima $(\pm 1,0)$ but not the saddle point $(0,0)$, and map that domain onto $R^2$ by a diffeomorphism. (The domain can be chosen of convenient starlike shape, so that the diffeomorphism can be given explicitly.)

EDIT on "properness" assumptions (for those that have "topology from a differential viewpoint" in their hands.)

Consider a ball $B_R$ centered at $(0,0)$ and of radius $R$. Let $V$ be the gradient vector field of $f$. By the theorem on the indices of zeors of vector fields, we known that

1) if $V$ points outward $B_R$ at its boundary, then the sum of the indices of the zeroes of $V$ equals the characteristic of $B_R$ which is $1$. Since minima contribute with $+1$, if $f$ has two minima it must have at least another critical point with index $-1$ (for instance a saddle).

2) if $V$ does not points outward $B_R$ at its boundary, then the sum of indices is the degree of the map $\partial B_R\to S^1$ given by $V/||V||$.

So if you assume that there is $R$ so that $V$ points outward $B_R$ at its boundary, then $f$ must have another critical point. On the other hand this tells you that with the above construction of an $f$ with only two minima, there is no $R$ such that the two minima are contained in $B_R$ AND $V=grad(f)$ points outward $B_R$ at its boundary.

In conclusion, everything depends on the degree of $V/||V||$ on big circles.

Also, you could replace $B_R$ with level sets. Suppose there is a simple closed curve $C$, contained in a level set (of a regular value) for $f$. By Jordan theorem $C$ bounds a disc $D$. Moreover, since $C$ is in a level set, the gradient vector field $V$ points always either inward or outward $D$ at $C$.

Thus the sum of indices of critical points in $D$ is $1$. So,

Fact: $D$ contains at least a critical point and if it contains two minima, it must contains also a third critical point.

In particular, one can use that to prove 2) if one adds the assumption that $f$ is proper, that is to say, $f$-preimages of compact sets are compact. (which is equivalent to the requirement that $\lim_{p\to\infty}f(p)=\infty$.)

In this case, suppose such an $f$ exists, let $B$ a ball containing the two critical points and let $K=100\sup_Bf$. $K$ is a regular value of $f$ because $f$ has only the two minima in $B$. Thus $f^{-1}(K)$ is a $1$-submanifold of $\mathbb R^2$. It is compact because $f$ is proper. Let $C$ be a connected component of $f^{-1}(K)$. $C$ does not intersect $B$ because of the choice of $K$. On the other hand, $C$ bounds a disc $D$ (by Jordan theorem).

Then, by the above Fact, if $D$ contains $B$ than it must contains a third critical point, and if $D$ does not contain $B$ then it contains a critical point which is different from the two minima in $B$. Therefeore $f$ must have at least three critical points.

Your topology instinct is right. If you consider the portion of the graph of $f$ inside a big ball, you will have two minima. If you cap off the boundary of that portion, you will have a topological sphere, with $\chi =2$. The height function has a maximum and two minima, which would give $\chi=3$. (I'm assuming nondegenerate critical points here.)

I'm still pondering an elementary argument. It would be helpful to know what you've learned/proved in your course.