The one-word answer is torque.
The door is not moving. That means that all forces must balance. So far so good. That is where the vertical forces on the hinges come from. The door has weight, which is in your diagram. (Don't let anybody be snide about your diagram. It's better than I would have done free-hand.) You have to assume something about the stiffness of the door and hinges to get the fraction of the door's weight each hinge holds, but never mind that for now.
But the door is of finite size. And therefore, when you apply the weight of the door as net being at it's center of mass (COM) it winds up applying a torque relative to the hinges.
This is the thing about torque. It requires a center around which to calculate it. When you apply the definition $\overrightarrow{\tau} = \overrightarrow{r} \times \overrightarrow{F}$ you need an origin for the $\overrightarrow{r}$.
And here's the thing about picking an origin for torque. The door is not rotating around any possible origin. It's not moving at all. So the torque has to be zero around any possible origin.
So you can pick a helpful one. For example, you could pick the top left corner of the door. Or the middle-left edge. And then you sum up all the torque values around each of those points, and set it to zero.
Now notice, the door is not moving left or right. So the horizontal force on one hinge has to be the negative of that on the other hinge, because those are the only horizontal forces, and they have to cancel.
And pretty quickly you see where the forces come from. And, hopefully, you will see why if you tried to support yourself off a door, the hinge that most probably fails first is the top one.