The answer is 'yes' when $G$ is compact. (*More generally, if the $G$-stabilizer of $m\in M$ is a compact group, $K\subset G$, then $\pi'(m):T_mM\to T_[m](M/G)$ is surjective; see the remark below.*) Here is an outline of the argument:

Note that all of the $G$-orbits in $M$ are closed in $M$. For $m\in M$, let $U\subset M/G$ be any open neighborhood of $\pi(m)\in M/G$, then $\pi^{-1}(U)\subset M$ is a $G$-invariant open neighborhood of the $G$-orbit of $M$, and we can replace $M$ by this open set without changing the problem (or any of the hypotheses). Thus, we can assume that $M/G\simeq \mathbb{R}^q$ for some $q>0$, and that $\pi = (x_1,\ldots, x_q)$ for smooth $G$-invariant functions $x_1,\ldots x_q$ on $M$ that vanish at $m\in M$.

Next, by the Slice Theorem, by shrinking to a smaller $G$-invariant neighborhood of $M$ if necessary and fixing a $G$-invariant metric on $M$, we can exponentiate the normal subspace $W\subset T_mM$ to the tangent space of the $G$-orbit of $m$. Then, letting $K\subset G$ be the closed subgroup that fixes $m$, we can reduce the question to the case of the compact group $K$ acting linearly on the $K$-representation $W$ and $\pi:W\to\mathbb{R}^q$.

Let $W = W_0\oplus V$ where $W_0$ is a trivial representation of $K$ and $V$ is a $K$-representation with no trivial summand. We can factor out the $W_0$ from the problem and reduce to the case of $K$ acting on $V$ without any trivial representation.

Now, I claim that this situation *never* satisfies Conditions 1 and 2, with $V/K$ being a topological manifold with a smooth structure such that $\pi=(x_1,\ldots,x_q):V\to V/K=\mathbb{R}^q$ is smooth.

The reason is as follows: The ring $R$ of $K$-invariant polynomials on $V$ is finitely generated (Hilbert) in degrees greater than or equal to $2$. Since $K$ is compact, there is at least one quadratic $K$-invariant polynomial, say, $Q$, that is positive definite. By Condition $2$, there is a smooth function $F:\mathbb{R}^q\to\mathbb{R}$ satisfying $F(0)=0$ but $F(x)>0$ for $x\not=0$ such that
$$
Q = F(x_1,\ldots, x_q).
$$
Let us write $F$ in the form
$$
Q = F = c_1\,x_1+\cdots + c_q\,x_q + H(x_1,\ldots,x_q)
$$
for constants $c_1,\ldots,c_q$ and where $H$ vanishes to order at least $2$ at $0\in\mathbb{R}^q$. Now, since there are no nonzero $K$-invariant linear functions on $V$, all of the $x_i$ vanish to order at least $2$ at $0\in V$, and hence $H(x_1,\ldots,x_q)$ must vanish to order at least $4$ at $0\in V$. Consequently, the $c_i$ cannot all be zero since $Q$ is a positive definite quadratic form on $V$.
Thus, the smooth function $F:\mathbb{R}^q\to\mathbb{R}$ has a nonvanishing first derivative at $0\in\mathbb{R}^q$ while $0$ is a strict local minimum of $F$, which is impossible.

Thus, we must have $V=0$, i.e, $K$ must act trivially on $W$, and in this case, $M/K = W$, so Conditions 1 and 2 are fullfilled and, indeed, $\pi$ is a submersion.

**Remark 1:** The above argument basically relies on the Slice Theorem to reduce to the case of a smooth surjective map $\pi:T_m/T_m(G{\cdot} m)\to\mathbb{R}^q$ whose fibers are the orbits of the closed subgroup $K\subset G$ that fixes $m\in M$, and then uses the compactness of $K$ to show that Conditions $1$ and $2$ imply that the action of $K$ on the vector space $T_m/T_m(G{\cdot} m)$ must be trivial. Thus, the only place that compactness is needed is that $K$, the $G$-stabilizer of $m\in M$ should be compact.

Returning to the more general case, it's clear that, if the answer is going to be 'yes' for general $G$-actions, then the answer would at least have to be 'yes' for the simple case of $G$ represented faithfully on a vector space $V$ with the property that the space of orbits $V/G$ has the structure of a smooth manifold in such a way that the quotient map $\pi:V\to V/G$ is smooth and such that every $G$-invariant smooth function on $V$ is the $\pi$-pullback of a smooth function on $V/G$.