One way to see a rigid body is a collection of infinity particles such that they preserve the distances between themselves (can only translate and rotate). So, we have a constraint, and then internal forces that maintain this constraint.
This internal forces are dependent on the forces that you apply in the system. If you apply a force in one point of the object (one particle), then all other points would feel a force such that the accelerations of the hole system don't destroy the constraint. Rigid bodies need to have energy stored there to do that.
If you have a static homogeneous rigid body with the center of mass $\vec{r}_{cm}=\vec{0}$, applying a force $\vec{f}(\vec{r}_0)$ in some point $\vec{r}_0$ such that the force is paralel to $\vec{r}_0$, then all the points of the body need to feel this same force to preserve the constraint. The result of this force is an aceleration of the whole system generating a translation.
If the force $\vec{f}(\vec{r}_0)$ is not parallel to $\vec{r}_0$, then all the points need to feel a force $\vec{f}(\vec{r})$ such that
$$
\vec{r}\times\vec{f}(\vec{r})=\vec{r}_0\times\vec{f}(\vec{r}_0)
$$.
because the constraint is mathematically translated to $d\vec{r}=d\vec{\theta}_0\times\vec{r}$ for every $\vec{r}$ and the same $d\vec{\theta}_0$ (see here for understand why).
With this we can define a quantity called Torque associated for a force $f$ applied to a point $\vec{r}$ of the rigid body by:
$$
\vec{\tau}=\vec{r}\times\vec{f}(\vec{r})
$$
So, if we have a force $\vec{f}_1$ applied at $\vec{r}_1$ and $\vec{f}_2$ applied at $\vec{r}_2$ we can see that the net force at point $\vec{r}$ need to obey:
$$
\vec{f}(\vec{r})\times\vec{r}=\vec{f}_1\times\vec{r}_1+\vec{f}_2\times\vec{r}_2
$$
Then we may define a net torque:
$$
\vec{\tau}_{net}=\sum_{j}\vec{f}_{j}\times\vec{r}_{j}
$$
And, if $\vec{\tau}_{net}=\vec{0}$, then the body does not rotate because at each point $\vec{f}(\vec{r})\times\vec{r}=0$ and for $\vec{r}\neq \vec{0}=\vec{r}_{cm}$ we conclude that the force $\vec{f}(\vec{r})$ is paralell to $\vec{r}$.
does it also leads to ...
No, it doesn't.
A simple Counterexample:
Consider the figure below (the bar $\textrm {AB}$ and forces $F$ are on a plane parallel to $\textrm {xy}$ plane)
We have $\Sigma \vec F=\vec 0$, but, if we calculate vector sum of torques about point $\textrm A$ we will obtain $\Sigma \vec M_A=F (\overline{AB})\vec k\neq \vec 0$ ($\vec k$ is the unit vector of $\textrm z$ direction)
Best Answer
What the book meant by "two dimensions" is actually a system whose forces acting on are lying in a plane, i.e., they are coplanar forces. Since the torque, $$\vec M=\vec r\times\vec F,$$ is a vector product involving forces, it points in a direction perpendicular to that plane. If the forces are in the $xy$ plane, then the torque is in the $z$ direction.