because a body toppling would simply mean that the weight vector is acting in a direction towards a point on the surface outside the body ?
Well, the weight vector is always going to act straight down. But other forces on the block may cause it to fall.
What I didn't understand from your question was how the body began to topple. Was it pushed, did it fail mechanically, did the ground give way, ...?
I'm going to assume it was pushed. In the beginning, we just have a simple block sitting on the ground.
The weight vector points down and the supports of the block surround it, so there is no movement. In particular, let's consider the (marked) bottom right corner. Can the block pivot at this point and fall to the right? No. Even if we remove the ground under the block so that no normal force is present, the torque about the corner would make it fall to the left, not to the right.
Now we start pushing the block to the right with an additional force $F$.
Let's consider the torque exerted about the same corner on the right. We have the original counter-clockwise torque from the weight of the block ($mgx$), but now a clockwise torque from the pushing ($Fy$). If $Fy$ is greater than $mgx$, then the torque will be sufficient to accelerate the block around the corner and it will tip.
There is always a normal force holding up the block, but if it is tipping (or nearly tipping), then only the corner of the object can be supporting the weight. Therefore the normal force moves until it is acting through the corner.
With no distance to act, there is no torque from the force, so we can ignore it for calculating if the tipping will happen.
A lot of material in your question. Let me try to summarize:
You have an cube sliding down a frictionless slope of 45°, when it suddenly encounters a surface with friction $\mu$. Will it topple?
The answer is "yes" if the sum of two forces - gravity and deceleration - points along a line that does not pass through the base of the cube. Since the cube was on a 45 degree slope in your example, the line of action of the force of gravity is already pointing at the corner of the cube:
If you add any friction, that will result in an additional torque on the block that will topple it.
The problem is more interesting when you have a more stable situation - lower angle of slope (or wider block): in that case, the question of whether the block will topple (assuming that the coefficient of friction is unchanged with contact area - which is a reasonable assumption) has a more complex answer (but it's not what you asked for).
Best Answer
No your explanation is not missing anything.
Objects topple on inclined planes for the same reason they topple when pushed on horizontal planes.
If the object does not slide, the edge of the base will form a pivot about which the object can turn. Suppose the object is tilted gradually about this pivot then released. If the vertical line through the CG remains inside the base, then the weight forms a torque about this pivot which turns the object back onto its base. If this line falls outside the base, the torque due the the object's weight continues to turn the object away from its base.
Source : BBC Bitesize Physics website
If the object does slide, accelerating down the incline, then the situation is more complex. See BowlOfRed's answer to What causes a tall box to tip over on an inclined plane?