The pressure on each side of the pipe is given by $P = \rho g h$ where $\rho$ is the density, $g$ is the acceleration due to gravity and $h$ is the depth of the pipe.
From the Euler equations of motion in 1D and steady state, we have:
$$ \frac{1}{2}\frac{d u^2}{d x} = - \frac{1}{\rho} \frac{dP}{dx}$$
If we make some more assumptions, namely that the pipe is full of seawater (which makes sense in the steady state because the water will flow from the sea to the lake) and that the pressure at the fresh water end is constant (so we ignore dilution/mixing, the sea water just instantly drops under the fresh), and we integrate from $x = 0$ at the sea end to $x = L$ at the fresh end:
$$ u^2 |_0^L = - 2\frac{1}{\rho} P|_0^L$$
and taking the velocity to be zero at $x = 0$ gives:
$$ u(L)^2 = -2\frac{1}{\rho_s}(P(L)-P(0))$$
$$ u(L)^2 = -2\frac{1}{\rho_s}(\rho_f g h - \rho_s g h)$$
Taking $\rho_s = 1020 \text{kg}/\text{m}^3$ and $\rho_f = 1000 \text{kg}/\text{m}^3$ with $g = 9.8 \text{m}/\text{s}^2$ yields:
$$ u(L)^2 = 0.0961h $$
or
$$ u(L) \approx 0.31h^{1/2}$$
Obviously this makes some pretty big assumptions. No viscosity, which is probably not that bad of an assumption unless your pipe is really deep, and the pressure on the fresh water end is constant implying the salt water just "disappears" by dropping very quickly out of the pipe under the fresh water.
Why does the length of the pipe matter
It doesn't actually. You'll notice $L$ doesn't appear anywhere in the expression. The velocity at the end of a mile long pipe or a 1 inch long pipe is the same and given by that expression.
What is the significance of $h^{1/2}$
Again, there really isn't any significance. The units of pressure/density are $\text{m}^2/\text{s}^2$ which is what RHS of $ u(L)^2 = 0.0961h $ is. So the units on the 0.0961 are $\text{m}/\text{s}^2$ and $h$ is $\text{m}$. So when you take the square root of both sides to get into $\text{m}/\text{s}$, you end up with the $h^{1/2}$.
So the significance is really just that there is a non-linear relationship between velocity and the depth of the pipe. If you put your pipe four times deeper, you'll only get twice the velocity.
First, water at seabed pressure will also be between the sand grains in the bucket, so the water pressure does not add up to the force that will oppose your pull. This is already at equilibrium. So away go your 9800 N.
I'm not sure either about the 231 N, as the sand will probably mostly stay with the seabed as you pull out the bucket.
So your hope is in actual suction, that is, how much force is necessary to let water in when you pull the bucket out. First, as noted by User58220, this will not resist a continuous pull, as a flow of water will start as soon as there is any pull up. So unless the cylinder is denser than water and expels it slowly after a pull, it will eventually come out after some number of pulls.
If you assume that the boundary condition is really that the sand around the cylinder wall is as packed as the rest of the seabed, then you need to estimate the permeability $k$ of the sand there and from that apply D'Arcy's law to get the flow rate as a function of force, wikipedia, with the pressure drop equal to the pulling force divided by cylinder area. You're interested in $T=Ad/Q$, the time to pull out by a distance $d$ of the cylinder, as a function of the force, $T\simeq 2\mu Ahd /(kF)$. You can work out the time for it to settle back under its weight (this is true only if $d$ is small enough that the sand did not move).
What I fear is that you'll have a detachment between the packed sand and the walls of your cylinder, and flow will be much easier there. Cylinder should be very rigid and have rough walls at the scale of sand grains to try to prevent this (gluing sand on it is used in related experiments)
Best Answer
The pressure of the column of water between the two pistons will cancel out the pressure needed for the one down below to push the water up. So you are correct here.
I will have to correct you about one thing you mentioned in you question though. The pressure you'll have to overcome, should you pump from the lower pump directly is not all of the 10.9204 atm, since you already literally pump air from the surface through the air lines where the pressure is already 1 atm (neglecting the 100 mt air column-you might want to account for that). That pressure aids in your pumping. Even if these lines weren't there you would still be pumping at sea level, so either case you don't account for the atm. pressure.
The forces you will need to really over come when pumping from the surface is the friction in all the lines.