Suppose that I'm alone on a planet of $M$. From what I learned from school, the gravitation force acting on me is given by
$$F=G\frac{mM}{r^2},$$
where $m$ is my mass, $r$ is the distance between me and the planet and $G = 6.67 \cdot 10 ^{-11}$.
Suppose that my dog, of mass $\tilde m$, comes to live with me on this planet. Now, is the gravitation force acting on me still $F$ or is it
$$\tilde F=G\frac{m(M+\tilde m)}{r^2}~? $$
The vertically downward gravitational force exerted by the Earth of mass $M$ on a object of mass $m$ standing on the surface of the Earth is given by:
$F=G\frac{m \times M}{r^2}$ where $r$ is the radius of the Earth.
Now add a dog standing close to you, also on the surface of the Earth, then the dog would exert a (negligible) more or less horizontal force of attraction on you given by:
$F'=G\frac{m \times \tilde m}{d^2}$, where $d$ is the distance between you and the dog.
Since as $F$ and $F'$ are more or less perpendicular to each other (one vertical, one horizontal) the dog does not affect $F$ in the way you describe and:
$\tilde F=G\frac{m(M+\tilde m)}{r^2}$ is not correct. See also the force diagram below:
One could calculate a resultant force $F_R$ by vector addition of $F$ and $F'$. It would have the magnitude (scalar):
$F_R=\sqrt{F^2+F'^2}$ and with $F \gg F'$, then basically $F_R=F$.
I mean, technically it would affect F, yes. But the mass of your dog is negligible compared to the mass of the planet. Just going off of Earth, we're talking 6e24 kg compared to 50 kg at most. The gravitational force from the planet is so much more than the gravitational force from the dog - which is why in real life we fall towards the Earth, not towards our pets.
EDIT: The effect of the dog would be horizontal, not vertical as Earth's gravitational force is. So OP's (M+m) isn't right, but that doesn't change that we can ignore the dog's mass when calculating gravity.
