Plotting a contour on a torus

You could also use MeshFunctions option to map the $[0,1]^2$ region:

yourFunc = Function[{u, v},
                    Re[2 Exp[2 π I (u + 2 v)] + 3 Exp[2 π I (u - 2 v)]]
                   ];

ParametricPlot3D[{
                  (2 + Cos[2 π v]) Sin[2 π u],
                  (2 + Cos[2 π v]) Cos[2 π u],
                  Sin[2 π v]},
                 {u, 0, 1}, {v, 0, 1},
       MeshFunctions -> Function[{x, y, z, u, v}, yourFunc[u, v]],
       Mesh -> {{0}}, (* Because you state yourFunc[u,v] = 0 *)
       MeshStyle -> Directive[Blue, Thick],
       PlotPoints -> 50
 ]

Another fancy example:

ParametricPlot3D[{
                  (3 + Cos[2 π v]) Sin[2 π u],
                  (3 + Cos[2 π v]) Cos[2 π u],
                  Sin[2 π v]},
                 {u, 0, 1}, {v, 0, 1},
          MeshFunctions -> Function[{x, y, z, u, v}, yourFunc[v, u]],
          Mesh -> {Range[-1, 1, .1]},
          MeshStyle -> None,
          MeshShading -> Join[{None},
            ColorData["Rainbow"] /@ Rescale[Most@Range[-1, 1, .1]],
            {None}],
          PlotPoints -> 100,
          PlotStyle -> None,
          Lighting -> "Neutral"
 ]

One easy way is to use the Line's from your parametric curves and map them onto a torus mathematically. What basically happens here is that we interpret the points of your plot no longer as Cartesian coordinates but as angles $\phi$ and $\theta$ of the torus parametrisation. Here the function for this parametrisation:

With[{rr = 3, r = 1},
 torus[{u_, v_}] := {(rr + r*Cos[2 Pi u])*Cos[2 Pi v], 
   (rr + r*Cos[2 Pi u])*Sin[2 Pi v], r*Sin[2 Pi u]}
]

rr is the inner radius and r the radius of the tube. Now let's take your plot and do the transformation. In the following code, all the magic happens at where I extract the lines of gr using Cases and then transform them into Tube's using our torus function.

gr = ContourPlot[Re[2 Exp[2 Pi I (x + 2 y)] + 3 Exp[2 Pi I (x - 2 y)]] == 0, 
  {x, 0, 1}, {y, 0, 1}];
Show[
 ParametricPlot3D[torus[{u, v}], {u, 0, 1}, {v, 0, 1}, 
  PlotStyle -> Directive[Opacity[.3], Red], Mesh -> None],
 Graphics3D[{Blue, 
   Cases[Normal[gr], Line[__], Infinity] /. 
    Line[pts_] :> Tube[torus /@ pts]}]
]

For some applications, it might be useful not to use the conventional parametrization of the torus. In particular, if one wants a conformal (angle-preserving) map from a rectangular domain to a torus, one can use the parametric equations (see e.g. Sullivan's paper):

$$\mathbf r(u,v)=\frac1{\sqrt{s^2+t^2}-t\cos\tfrac{2\pi v}{t}}\begin{pmatrix}s\cos\tfrac{2\pi u}{s}\\s\sin\tfrac{2\pi u}{s}\\t\sin\tfrac{2\pi v}{t}\end{pmatrix}$$

with parameter ranges $0\leq u\leq s,\; 0\leq v\leq t$ to conformally map an $s\times t$ rectangle onto a torus.

In particular, for your curve of interest, we can do something similar to Silvia's proposal, but using this parametrization instead. If you want to see the curve only, without the torus backdrop, you can do this:

With[{s = 1, t = 1}, 
     ParametricPlot3D[{s Cos[2 π u/s], s Sin[2 π u/s], t Sin[2 π v/t]}/
                       (Sqrt[s^2 + t^2] - t Cos[2 π v/t]),
                      {u, 0, s}, {v, 0, t}, Mesh -> {{0}},
                      MeshFunctions -> Function[{x, y, z, u, v},
                                      Re[2 Exp[2 π I (u + 2 v)] + 3 Exp[2 π I (u - 2 v)]]],
                      MeshStyle -> Directive[Thick, Blue],
                      PlotPoints -> 55, PlotStyle -> None]]

(Change s and t if you want a plot over a wider rectangle.)

If you want to see the torus as well, you can have it be translucent, so that you can still see the structure of the curve. For instance, here's the resulting picture with the setting PlotStyle -> Opacity[3/5, ColorData["Legacy", "PowderBlue"]]: