Nodal voltage analysis in a DC circuit
You have done a good job drawing arrows to represent your currents.  When you apply Kirchoff's current law you are just saying all current into a node = all current out of a node.
Below I'll write the node A equation and will arbitrarily assign current into the node as positive:
$$
 \frac{V_{G_1}-V_A}{R_1} - \frac{V_A-V_{reference}}{R_2} - \frac{V_A-V_{reference}}{R_3} - \frac{V_A-V_B}{R_4} - I_{G_2} = 0
$$
Equally as valid - i can arbitrarily assign current into the node as negative which would result in the following equation:
$$
 -\frac{V_{G_1}-V_A}{R_1} + \frac{V_A-V_{reference}}{R_2} + \frac{V_A-V_{reference}}{R_3} + \frac{V_A-V_B}{R_4} + I_{G_2} = 0
$$
Clearly, these are both equivalent.
EDIT:  Using your figure and adding an arrow for IG2.  I will arbitrarily decide that current into node B is negative and write a KCL equation:
$$-I_{R_4} - I_{G_2} - I_{R_5} - I_{G_3} = 0$$