Draw two vectors [itex]x,y[/itex] on a copy of [itex]\mathbb R^2[/itex]. Or on a piece of paper, if that's all you have available.
Draw an arrow from [itex]x[/itex] to [itex]y[/itex] on your paper. Its tail should be at [itex]x[/itex] and its head at [itex]y[/itex]. We can think of the arrow as the vector [itex]z:= y-x[/itex]. Indeed, if we moved the arrow so that its tail is at the origin, the head would lie at [itex]z[/itex].
-Imagine an ant is sitting at [itex]x[/itex] and you want it to travel to [itex]y[/itex] in a straight line. What should it do? It should travel all the way along [itex]z[/itex]. This would bring it to [itex]x + z[/itex], also known as [itex]y[/itex].
-Now, imagine the ant is sitting at [itex]x[/itex] again, and you want it to travel [itex]\frac13[/itex] of the way to [itex]y[/itex] in a straight line. What should it do now? It should travel [itex]\frac13[/itex] of the way along [itex]z[/itex]. This would bring it to [itex]x + \frac13z[/itex].
-And what if you want it to travel (again, from [itex]x[/itex]) [itex]\frac57[/itex] of the way to [itex]y[/itex] in a straight line? Now, it should go to [itex]x + \frac57z[/itex]
This is the sense in which [itex]\{x+tz: t\in[0,1]\}[/itex] parametrizes the line segment between [itex]x[/itex] and [itex]y[/itex]. Finally, notice that [itex]x+tz = (1-t)x + ty[/itex].