The first (convex) lens produces an image that is to the right of the diverging lens i.e. this acts as a virtual object for the diverging lens. So the rays look like the diagram below. I've drawn a point object to keep the diagram simple. This could for example be an image of a distant star.
![Virtual object](https://i.stack.imgur.com/nttm8.gif)
When we say there is a virtual object we mean that to the left of the lens the light rays are converging as if they were coming to a focus at the point where the virtual object is. I've drawn those converging rays as solid blur lines to the left of the lens and as dashed line to the right of the lens to show how they would come to a focus at the object if the diverging lens was not there.
Now the diverging lens makes the rays diverge, which in this case means it reduces their convergence. With the diverging lens in place the light rays look like this:
![Diverging lens](https://i.stack.imgur.com/HHwm3.gif)
The diverging lens refracts the converging rays to be parallel i.e. as if they were coming from an object at infinity. This is how the diverging lens takes a virtual object at the focal point and produces a virtual image at infinity. The lens in your eye then brings the parallel rays to a focus on your retina so you can see the image.
Your diagram is actually perfectly correct, but it doesn't show what is happening in the telescope. Your diagram shows a virtual object at $u = f/2$ forming a real image at $v = f$, or by reversing the rays a real object at $u = f$ forming a virtual image at $v = f/2$.
We'll use the Cartesian convention, and to avoid possible sign confusions I'll write the focal length of the lens as $f = -F$, where $F$ is a positive constant. Then if we consider a virtual object a distance $F/2$ to the right of the lens that is at $u = +F/2$. Feeding this into the lens equation:
$$ \frac1u + \frac 1f = \frac1v $$
We get:
$$ \frac2F + \frac{-1}{F} = \frac1v $$
So $v = +F$ i.e. a real image at a distance $F$ to the right of the lens. If we reverse the rays we get a real object at a distance $F$ to the left of the lens, i.e. $u = -F$, so:
$$ \frac{-1}{F} + \frac{-1}{F} = \frac1v $$
Giving $v = -F/2$ i.e. a virtual image at a distance $F/2$ to the left of the lens. Neither of these match the situation in the telescope where we start with a virtual object a distance $F$ to the right of the lens i.e. $u = +F$. Putting this into our equation we get:
$$ \frac{+1}{F} + \frac{-1}{F} = \frac1v $$
so $1/v = 0$ i.e. the image is at infinity.
The reason your diagram gives the wrong result is that the direction of the light rays defines the positive direction. In your first diagram the light rays travel left to right, which is the usual convention, so positive is to the right. In your second diagram the (virtual) object is to the right of the diverging lens, so the (virtual) light rays have to be travelling towards the object i.e. left to right. You have drawn the rays travelling right to left, and that makes your object a real object not a virtual one.
Drawing the diagram for the (virtual) object at $u = +F$ and the (virtual) image at $v = -\infty$ is a bit hard, so to illustrate what the diagram looks like I've put the (virtual) object at $u = +\tfrac32 F$. This creates a (virtual) image at $v = -3F$:
![Virtual object](https://i.stack.imgur.com/RXNWb.gif)
Note that all light rays, real and virtual, travel left to right. If you move the (virtual) object leftwards towards $F$ the (virtual) image move leftwards towards negative infinity.
Best Answer
It is easier to think about this in reverse - i.e. what does it mean for light to be approaching from infinity? When we reference light from an object at infinity, we mean that the object is so far away that all light rays (strictly speaking in terms of geometric optics) from the object appear as if they are parallel and traveling in the same direction. When these parallel rays approach the convex lens, they get focused to the focal point of the lens.
In optics, the direction of light travel does not matter, thus when an object is placed at the focal point of the lens, the light rays will become such that they are perfectly parallel and traveling in the same direction. This is usually called "collimated" light.
Collimated Light
This method of collimating a light source (or object) is generally used in various imaging techniques such as shadowgraph and Schlieren.
Now what will happen if you place your hand in the optical path of the light? Well, theoretically, you should see nothing since no image is formed from parallel light. But in reality the light will never be perfectly collimated so you may see a very blurry version of whatever your object is. As soon as you place another focusing element in its path though, all those light rays will be focused at its focal point and you will (theoretically) have a perfectly formed image of your object.