Characteristic spectra 
Why are the "K-lines" so much higher than the bremsstrahlung line?
Are loads of x-rays emitted at the characteristic wave-length, or is the energy of each photon much higher?
Why isn't this a dot, or vertical line, but a sharply gradated curve?
I've been unable to conclusively find out why.
How I interpret the graph:
The amount of low/high energy x-ray photons emitted is low.
Near the middle, the number is higher, thus the intensity/count is higher(?)
The K-lines represent one photon emitted from when an electron drops, but because it's energy is so much higher than the sum of the bremsstrahlung radiation, it spikes up.
$$
I = \frac{P}A \propto \frac{E}T \propto Nf \propto \frac{N}\lambda
$$
where N is the number of photons, and lambda is the wavelength.
So for the intensity to be so high, wouldn't the Number of emitted photons increase? The wavelength is specified on the graph
 A: The answer by Jaromrax is fine with the physics. Now trying to see your confusion:

Characteristic X-rays are emitted when outer-shell electrons fill a vacancy in the inner shell of an atom, releasing X-rays in a pattern that is "characteristic" to each element. Characteristic X-rays were discovered by Charles Glover Barkla in 1909,1 who later won the Nobel Prize in Physics for his discovery in 1917.
Characteristic X-rays are produced when an element is bombarded with high-energy particles, which can be photons, electrons or ions (such as protons). When the incident particle strikes a bound electron (the target electron) in an atom, the target electron is ejected from the inner shell of the atom. After the electron has been ejected, the atom is left with a vacant energy level, also known as a core hole. Outer-shell electrons then fall into the inner shell, emitting quantized photons with an energy level equivalent to the energy difference between the higher and lower states.

Since the target is bombarded by electrons, the brehmstrahlung spectrum will appear as a continuous line, from electrons interacting with the fields in the lattice. That spectrum is continuous. BUT electrons can hit and eject an  inner electron. The rest of the electrons in the atom will fall in ejecting a specific frequency K photon which will add up to the Brehms spectrum.
with this in mind let us see your confusion:

Why are the "K-lines" so much higher than the bremsstrahlung line? Are loads of x-rays emitted at the characteristic wave-length, or is the energy of each photon much higher?

Loads of electrons hit an orbital with a K line , and the hole left is immediately filled by a decay of a higher orbital electron emitting the characteristic  Xray. The energy of a photon= h*nu, so it is a fixed energy at each lamda. The number is large. Higher beam intensity, more K hits and transitions filling the hole. These add up on the number of hits due to photons from brehms .

Why isn't this a dot,

This is a histogram, hits added on a bin. A dot would mean only one hit on the detector, and would not be visible within statistical errors.

or vertical line,

because of thermal motions and accuracies of detectors the inherent width of the transition becomes a gaussian , as discussed in the other answer.
A: Bremsstrahlung originates from slowing down the electrons (in this case). They have different velocities and irradiate different energies with some distribution (see http://journals.aps.org/pr/pdf/10.1103/PhysRev.77.165).
K-line originates from a transition in a discreete energy (see https://en.wikipedia.org/wiki/K-alpha). The process itself has an energy fluctuation so small that you cannot see it with your detector. But why dont you see a thin line in spectra? Because of the noise in your detector - the detector has a limited resolution due to (thermal, electrical and whatever other) noise. So - instead of one point - you see a Gaussian shape.

The area of the peaks (or bumps) in the spectrum tells us about the intensity of the effect. A peak may be very high, but its area (proportional to number of events) is smaller comparing to the wide bump.
Edit: The spectra (histogram) is recorded on event by event basis. Sometimes it can happen that 2 photons enter at the same time, but for this spectrum, you can (if brehmstrahlung is not special in this sense) forget it. The spectrum now looks similar to Compton + peaks, so I dont want to comment on brehmstrahlung anymore.
If the photon from a certain process hits the detector with an energy $E_i$, one can be lucky and record this energy (plus some $\Delta E$ from noises)  in the spectrum (histogram). At this moment, the histogram bin (channel) content (with the energy range around the total $E_i + \Delta E_i$) is increased by 1. Thickness of the K-line peaks corresponds to this $\Delta E$. Their area corresponds to how frequently the process happens. Hope this get us on the same terminolgy grounds
