It's a pentode (with the suppressor tied to the cathode voltage.) And I'd like to avoid discussing it and instead stick with operating it as a triode -- grid 1 only. The datasheet has a simple chart for that (admittedly with the screen set to a nominal voltage) and, besides, your questions make it clear this is a good place to start.
Suppose you were expecting to operate this with a \$B^+\$ voltage of \$150\:\text{V}\$ and were using a plate resistor of \$20\:\text{k}\Omega\$. I've placed that load-line in blue:
I've selected a quiescent plate voltage of \$100\:\text{V}\$, allowing a swing of \$\pm 10\:\text{V}\$, as shown above with the thick green line along the x-axis. The quiescent plate current is \$2.5\:\text{mA}\$.
In rough terms, it appears that the grid voltage swings from \$-600\:\text{mV}\$ to about \$-1.9\:\text{V}\$, with a quiescent point of about \$-1.2\:\text{V}\$. The quiescent point isn't exactly in the middle so this suggests some distortion over the range. But not a lot.
The gain will be about \$\frac{90\:\text{V}-110\:\text{V}}{-600\:\text{mV}-\left(-1.9\:\text{V}\right)}\approx -15\frac13\$ and \$g_m\approx 770\:\mu\text{S}\$.
These are often set up with a grid leak resistor galvanically tying the grid to ground and a cathode resistor that, with the quiescent current applied to it, jacks up the cathode a little bit to achieve the relatively negative grid voltage. Here, it suggests that \$\frac{1.2\:\text{V}}{2.5\:\text{mA}}=480\:\Omega\$. So a \$470\:\Omega\$ resistor would likely be selected. But keep in mind that different tubes of the same marking may settle differently.
The Characteristics and Typical Operation specifies a plate voltage \$90\:\text{V}\$ and a quiescent current of \$3.5\:\text{mA}\$, saying \$g_m\approx 900\:\mu\text{S}\$ (using the same screen voltage indicated in the chart.) This is consistent with what was just computed for a slightly different operating point.
With all that out of the way, the reason that the grid voltage is usually (almost always) operated negative with respect to the cathode is for about the same reason that a JFET's gate is operated so that it is reverse-biased with respect to its source. If the grid were positive, it would attract electrons from the cathode and capture many more than if it repelled them. Such a grid current is not desired.
Just think of it as a JFET, more or less, except for the power supply rails.
If you look back at the above curves and imagine a small plate voltage of, say, \$15\:\text{V}\$, you'd wind up in all that mess with the curves diving back towards the origin. There's little good to be had there in the Ohmic region.
