<p>I want to make an order with this configuration</p>

<ul>
<li>Arduino MEGA 2650 R3</li>
<li>Ramps 1.6 Plus</li>
<li>2 TMC2130 </li>
<li>2 Stepper motors 17hs3401</li>
<li>1 Fan</li>
</ul>

<p><img src="https://i.stack.imgur.com/Nl0sE.png" alt="stepper_motor"></p>

<p>But I am not sure what voltage and current would be enough to make it work. The Ramps 1.6+ board has an input of 12V as you can see in the picture, but I have read that <a href="https://3dprinting.stackexchange.com/questions/9911/tmc2130-external-vm-ramps1-4">other people</a> had to increase the voltage with a DC-DC converter.</p>

<p><img src="https://i.stack.imgur.com/RtBEr.jpg" alt="ramps_voltage"></p>

<p>So, would be 12V and 15A enough to make work that configuration?</p>

The easiest way to know how powerful the PSU should be is to download from
https://github.com/rcarlyle/StepperSim
the Excel workbook which simulates the power absorbed by the stepper motor. Input the motor specifications, check in the graph the max speed at which you plan to run it, check the absorbed power, add 20% for the various losses. Once you know the absorbed power, you can multiply by the number of stepper motors and obtain the current required, add 20% not to stress the PSU too much.
In you case, this is the result
Your motor will never absorb more than 7 W, 14 W for the two motors, around 20 W considering losses in the motors and overcapacity of the PSU. This means that around 2 A PSU will be perfect for your setup.
Previous answer, which may not be completely correct but still provides a useful background knowledge:
The TMC drivers limit the current to the value you set, which is most of the time lower than the datasheet. In no case you will need more than 1.3 A * sqrt(2) * 2 motors = 3.6 A. In fact, you won't need this much either.
When there is no field in a coil, the driver applies full voltage, but the current is low (initially zero) so you don't hit the 1.3 A per coil.
The current increases (the higher the maximum voltage available to the driver, the faster it increases) and the driver (probably) keeps the full voltage until the preset current is reached. Just a moment before that, the current is almost there, but you still have full voltage from the power supply. This is the theoretical worst case, but it applies only for a very short amount of time.
As soon as the current reaches the preset, the driver "cuts the voltage" to keep  2.4 ohm * 1.3 A = 3.2 V (because V = R * I). This means that the power supply sees 3.2 A/12 V*1.3 A = 0.35 A.
When running, the motors almost never start from zero to max current: both coils are powered and when one increases, the other one decreases.
In fact, the microstepping makes the steppers act more or less like AC motors with two phases. This means that overall the current is the max current per phase multipled by sqrt(2). Also, when using microstepping one phase (coil) is not completely shut off, but two of them work at the same time (with different current levels). This means that in total one compensates the other, and the power supply only provides, more or less, 0.35 * 1.4 = 0.5 A per stepper. You have two, so it's 1 A total, therefore 2 A PSU considering the inefficiencies.
A very easy and complete explanation is provided
here
:
By controlling the duty cycle of the chopper, an average voltage and
  an average current equal to the nominal motor voltage and current are
  created.
...
As the current increases, a voltage develops across the
  sensing resistor, which is fed back to the comparator. At the
  predetermined level, defined by the voltage at the reference input,
  the comparator resets the flipflop, which turns off the output
  transistor. The current decreases until the clock oscillator triggers
  the flip-flops, which turns on the output transistor again, and the
  cycle is repeated
So you never have coming out of the PSU more than the preset current.
Supply current is not the same as the motor current in a copper drive
. It is the motor current multiplied by the dutycycle, at
  standstill typically Isupply = IM · ( VM ⁄ Vsupply )
...
Depending on
  how the H-bridge is switched during the turn-off period, the current
  will either recirculate through one transistor and one diode (path 2),
  giving the slow current decay, or recirculate back through the power
  supply (path 3). The advantage of
feeding the power back to the
  power supply
is the fast current decay and the ability to quickly
  reduce to a lower current level.