How fast is an equivalent 2-tape TM compared to a $O(n^2)$ 1-tape TM?
Just an extended comment to underline how the question is (up to my knowledge) far from being solved (and easy).
First of all there are no &quot;natural&quot; quadratic lower bounds with respect to multi-tape Turing machines (see e.g. K.W.Regan, On Superlinear Lower Bounds in Complexity Theory). So the approach, find an $O(n^2)$ problem on a 1-tape Turing machine that cannot be solved faster on a 2-tape TM would imply a major breakthrough.
On the opposite direction, one should be able to find a &quot;universal&quot; way to speed-up the simulation of a 1-tape TM on a 2-tape TM, but there are some examples for which the task seems hard; for example a 1-tape TM can efficiently simulate a dynamic non-linear system such as 1-D cellular automata - CA - (steps are parallel and use local information); and it is conjectured that the only way to know the state of the system after $n$ steps is to evolve the system $n$ times according to the rules (see e.g the (famous? :-) computational irreducibility - CIR - conjectured by S. Wolfram and analyzed in H. Zwirn and J.P. Delahaye, Unpredictability and Computational Irreducibility).
For example for every 1-D CA $A$ there is an $O(n^2)$ 1-tape TM $M_A$ that solves the problem:

Given a configuration $x$, is the state of $A$ equal to $x$ after $|x|$ steps from a blank initial configuration?

in $O(|x|^2)$ time.
Just use an expanded alphabet to overlap the target (input) configuration $x$, the current step evolving configuration and the next step configuration, and an extra bit for a &quot;distributed&quot; counter to stop the evolution after  $|x|$ steps.
E.g.
x = 1 1 0 1 0 0 1  (target configuration)

x = [1 0 0 s ] [100s] [000s] [100s] [000s] [000s] [100s]
     | | | +-step counter
     | | +---next step CA configuration
     | +-----current step CA configuration
     +-------target configuration

The TM sweeps from left to right and apply the CA rules writing the new configuration in the &quot;next step&quot; bits. At the end of a sweep the role of the next/current bits are swapped. Furthermore, at every sweep it deletes an $s$, when no more $s$  it simply compares the current configuration to the target configuration.
There are not known ways to speed-up the simulation of a 1D CA on a k-tape TM, and (probably) such speed-up would imply that the Computation Irreducibility conjecture is false.