Transconductance of a device
The transconductance tells you how much the current changes when you increase/decrease the gate/base a very tiny bit. It is a small-signal parameter. So a \$g_m = 25mS\$ at a \$v_{GS} = 0V, v_{DS} = 10V\$ like in your graph will mean that if you increase \$v_{GS}\$ a very tiny bit by \$\Delta v_{GS}\$, that the drain current will also increase a bit by \$\Delta i_d \approx 25mS \cdot \Delta v_{GS}\$.

For BJT's, the transconductance gain can be approximated by

$$g_m \approx \frac{I_c}{n V_T}$$

With \$n\$ the emission coefficient, \$V_T\$ the thermal voltage.

This means that the transconductance is proportional to \$I_c\$, or

$$g_m \sim I_c$$

For MOSFET's (similar to JFET's) the situation is a bit different. The approximation of the transconductance gain is here:

$$g_m \approx \frac{2 I_d}{V_{GS} - V_{TH}}$$

In order to make \$g_m\$ go up, we can just decrease \$V_{GS}-V_{TH}\$, however: the current \$I_d\$ will also decrease when doing that. It turns out that this decrease is approximately:

$$I_d \sim (V_{GS}-V_{TH})^2$$

So you can write that the transconductance is proportional to

$$g_m \sim \sqrt{I_d}$$

or

$$g_m \sim \frac{1}{\sqrt{V_{GS}-V_{TH}}}$$

And this is a bit annoying. This dependency is much slower! So in order to get the same \$g_m\$ for a FET, you will need a lot of current (limited by power consumption and velocity saturation, where the formula doesn't apply anymore), or almost no \$V_{GS}-V_{TH}\$ voltage (where \$I_d\$ will usually reach impractically low levels \$\sim nA\$). There is one way of solving this, and that is making the FET gigantic, but that is usually impractical as well and it makes other parasitic effects worse.