If the compression strength of iron is 50 MPa, or 5 MPa at 1000C, how can it be forged by a falling hammer, weighing say 10 kg, 10x10 cm impact area -> 100N/0.01 m2 = 10 kPa?
Having been a quasi-professional blacksmith for the past 13 years, I can provide a partial answer. Your calculation assumes that the pressure exerted by the hammer is just the weight of the hammer divided by the area of the hammer's face. In fact, the pressure should be calculated as the force due to deceleration of the hammer at the moment it strikes the workpiece, divided by the area of impact.
In order to get a sense of the deceleration, you can use the average depth of the impression left by a good hammer blow: around 2 mm. The speed of the hammer at the moment of impact is probably about the speed it would reach after falling freely for about 3 or 4 meters, or about 8 meters per second; and the hammer comes to a dead stop, or even rebounds, after traveling only a couple of millimeters subsequent to initial impact.
Also, a hammer blow is almost never perfectly flat. The hammer face is almost always is slightly tilted relative to the surface it hits, so all of the force due to that deceleration is initially concentrated in a few percent of the area of the hammer's face.
A seat-of-the-pants estimate, against which a properly calculated estimate should be compared, is that the hammer stops in 2 mm compared to a 4000 mm free fall, so the force exerted by the blow is on the order of 2000 times greater than the weight of the hammer. Add that to the slight tilt of the hammer face, and the force is probably more like 20,000 times greater than the weight of the hammer at the first moment of impact.
Another way to estimate the hammer's deceleration is to assume that the workpiece exerts a pressure equal to its yield strength, against the hammer. In the case of the slightly tilted hammer face, that pressure is initially exerted against only a small area so the deceleration is small; and the area increases as the impression size grows, so the deceleration increases. To calculate the deceleration vs time you would need to do an integral.