If there is no connection between A and B why are the resistors not in series, to me it looks like there is only one path?
They are in parallel because if you look at the circuit from AB, you have two paths: one passing through the 10 Ohms resistor, and the other passing through the 20 Ohms resistor.
As vicatcu pointed, they are in parallel because they share both terminals at the nodes A and B.
Secondly does shorting out the emfs mean removing them? Presumably if they where still connected the resistors would stop a short circuit.
Thévenin's analysis consists in synthesizing a linear circuit into a voltage supply and a resistor: being linear, the circuit will be characterized by a straight line, which will have a certain slope and offset. The slope is given by the equivalent resistance, and the offset is given by the equivalent voltage supply.
Mathemathically this coud be expressed as:
$$ V_{AB} = ( I_{AB} \cdot R_{th} ) + V_{th} $$
(considering Iab going into the Thévenin circuit)
Removing the voltage suppies means that you are analizing only the part of the circuit which depends by the current (and in this case it's represented by the resistors). Shorting them means "I remove all the offsets and consider just the slope"
It's like if you have a straight line passing for the point (0,2) in a graph: you can say that it's the sum of the same line passing in the origin (0,0) and the constant '2'.
The figure represents graphically the problem (click to enlarge):
on the x axis you have the current, and on the y you have the voltage. The Thévenin equivalent voltage source is a horizontal line, while the resistance determines the steepness of the total slope. The load resistor is a line passing for the origin, and they meet at the operating point.
Note that, arbitrarily fixed the direction of the current as going into the Thévenin equivalent, the resistor characteristic is inverted. You can also do the inverse.
Is the reason it is shown to go anti-clockwise because the 20V battery is more powerful than the 10V battery causing current to flow in that direction?
When you have a circuit to analyze, you first define a conventional sense for the current, and then you find the numerical value (positive or negative) with the corresponding sign. In this case, being a voltage source dominant, they choose that sense because it's more likely to reflect the real current flow.
Best Answer
That's a misunderstanding of the maximum power transfer theorem according to which the maximum power that can be delivered to the load is 50% of the maximum power available from the source.
But that doesn't mean that the best power ratio is 50%.
For a Thevenin voltage \$V_{th}\$, Thevenin resistance \$R_{th}\$ and load resistance \$R_L\$, the power delivered to the load is
$$P_L = \frac{V^2_{th}}{\left(R_{th} + R_L \right)}\frac{R_L}{\left(R_{th} + R_L \right)}$$
which is indeed maximum when \$R_L = R_{th}\$.
However, the power developed by the Thevenin source is
$$P_{th} = \frac{V^2_{th}}{\left(R_{th} + R_L \right)}$$
and so the fraction of the source power delivered to the load is thus
$$\frac{P_L}{P_{th}} = \frac{R_L}{\left(R_{th} + R_L \right)}$$
So, for \$R_L >> R_{th}\$, almost all of the power supplied by the source is delivered to the load (though this power is much less than the maximum power available from the source).
I don't have the expertise to address the reasons for 62 units of energy lost in the power plant so this answer is just to address the proper interpretation of the maximum power transfer theorem.