How does the resistor characteristic work on a MOFSET diagram?
The resistor characteristic (commonly called a "load line") simply reflects Ohm's Law.

The voltage across R is whatever is left over after we subtract the voltage drop across the MOSFET's channel from the supply voltage, and the current is simply I = V/R.

A potentially confusing aspect of the load line is that it is expressed in terms of the voltage on the device of interest, in this case the MOSFET, rather than the resistor's voltage.  This is why the line slopes downward instead of passing through the origin and sloping upward.

If the voltage across the transistor is almost 0V, then the resistor takes almost the entire supply voltage, and hence the current is 10 mA. Since that resistor is in series with the MOSFET's channel, the current through it is also 10 mA. The resistor current gives us the transistor current, according to Kirchhoff's Current Law.

Also, a device can indeed have nearly zero volts across it, and have current flowing. An ideal conductor exhibits no voltage drop.

To acquire good intuition about load lines, it is a good idea to first understand the load-line for a two-terminal semiconductor such as a diode (in series with a resistor).  A diode has a single I-V curve.  Where that curve intersects with the resistor load line is the operating point: this is very simple.

A transistor complicates the load line because it doesn't have a single curve. A transistor's graph is actually three-dimensional: for any given \$&lt;V_{DS}, I_{DS}&gt;\$ coordinate pair, there is \$V_{GS}\$ value. This 3D graph is folded into 2D by means of multiple contour lines which trace through the coordinate space for a fixed \$V_{GS}\$ value, much like elevation lines on a topological map, or isobaric lines on a weather map.  The simplest way to use the load line is to choose a \$V_{GS}\$ value. This reduces the 3D graph to a 2D graph: it establishes a particular curve to follow, labelled by that \$V_{GS}\$ value; and the other curves become irrelevant. Then, wherever that \$V_{GS}\$ curve crosses the load line is the operating point (for the given value of R corresponding to the load line!).

Note also that the graph could be even more complicated, because it could show multiple load lines, for different values of R. (Higher values of R will produce lines closer to the origin because higher resistancews limit the current more.) To "solve" for an operating point, we would have to then select the load line corresponding to our actual value of R, select a \$V_{GS}\$ curve, and read off the \$V_{DS}\$ and \$I_{DS}\$ values from the intersection of our selected load line and curve.