We do not know if the hanging weight will have a an instant of going downward or not
It depends on the mass and spring constant of the lower spring and mass of the hanging weight.
We cut the rope. There is gravitational force on the weight. The springs begin to contract rapidly and contract until the strings are tight. (John Hunter’s answer assumes the mass has to descend to make the strings taut, but the contracting springs do most of the moving to tighten the strings).
Assume the strings have no mass.
After cutting but before tight strings, the [lower spring / hanging weight] is a free system not affected by the upper spring or weightless un-taut strings. Let’s consider this system during that time:
The system as a whole is only acted on by gravity so as mentioned by alephzero it will fall - it’s center of mass will accelerate downward at F/M = g, the acceleration of gravity, where M = m_spr + m_hw (spring and hanging weight).
In addition to that, the spring (which has mass) will contract. As alephzero said, if the hanging weight has a much much higher mass than the lower spring, the mass will go down as this system falls (until the strings are tight).
But what if the spring doesn’t have a mass that is negligible compared to the hanging weight? There are cases where the spring constant and/or spring mass are high enough that the mass never (even instantaneously) has a downward acceleration of velocity. The spring contracts with enough inertial force that it’s instantaneous force on the weight is greater than m_hw g.
To be sure there are such cases, take an extreme case where the mass of the hanging weight is negligible compared to the spring’s mass and the spring constant is high. The spring will begin to fall under its own weight while contracting rapidly about its center and its lower end will go up (hence the hanging weight will)