The problem is that you have to consider the full $p_{system}$ when you look at $p_{system} (t_f)$.

The full momentum of the system has to consider every part of the system for it to remain conserved, and that includes the falling sand. So really you have:

$$p_{system} (t_f) = (m_{skater} + m_{sandbag} + m_{sand}) v_f$$

We can assume the falling sand has the same velocity $v_f$ because it should retain the velocity from when it fell (neglecting slight air drag), and since the surface is frictionless it should keep sliding along. If the sand were stopped by friction or air drag, that momentum from the sand would go into the ground or the air respectively; but wouldn't take momentum away from the skater or sandbag, or add any momentum to them.

If the sand were being pushed backwards relative to the skater when it fell out (not like here), then the velocity of the skater would increase; and there would be a force due to the sand pushing away from the skater.

The problem is that you have to consider the full $p_{system}$ when you look at $p_{system} (t_f)$.

The full momentum of the system has to consider every part of the system for it to remain conserved, and that includes the falling sand. So really you have:

$$p_{system} (t_f) = (m_{skater} + m_{sandbag} + m_{sand}) v_f$$

We can assume the falling sand has the same velocity $v_f$ because it should retain the velocity from when it fell (neglecting slight air drag), and since the surface is frictionless it should keep sliding along. If the sand were slowed by friction or air drag, that momentum from the sand would go into the ground or the air respectively; but wouldn't take momentum away from the skater or sandbag, or add any momentum to them.

If the sand were being pushed backwards relative to the skater when it fell out (not like here), then the velocity of the skater would increase; and there would be a force due to the sand pushing away from the skater.