Compute the Jacobians of the two mappings. I'll call your first mapping $X(u, v)$ so as not to re-use the variable $x$. Let $R > r > 0$.
$\frac{\partial X}{\partial(u, v)} = \left[ \begin{array}{cc} -\sin u & 0 \\ \cos u & 0 \\ 0 & -\sin v \\ 0 & \cos v \end{array} \right] $
$\frac{\partial f}{\partial(x, y, z, w)} = \left[ \begin{array}{cccc} rz & 0 & R + rx & 0 \\ rw & 0 & 0 & R + rx \\ 0 & r & 0 & 0 \end{array} \right] $
The embedding in $\mathbb{R}^3$ is the composite $f \circ X$, whose Jacobian is the product of the Jacobians:
$ \frac{\partial f}{\partial(x,y,z,w)}\frac{\partial X}{\partial(u,v)} = \left[ \begin{array}{cc} -rz\sin u & -(R + rx)\sin v \\ -rw\sin u & (R+rx)\cos v \\ r\cos u & 0 \end{array} \right] = \left[ \begin{array}{cc} -r\cos v\sin u & -(R + r\cos u)\sin v \\ -r\sin v\sin u & (R+r\cos u)\cos v \\ r\cos u & 0 \end{array} \right] $
(I used sage notebook www.sagenb.org for my computations.)
Now all of the questions concerning 1-to-1 can be phrased in terms of rank and nullity of these matrices. For example, it easy to see the first and third matrices have rank 2 (so nullity=0) for all choices of $u, v$, (which shows both representations of the torus are immersive, meaning an appropriate restriction of domain will make those maps 1-to-1).
Now $f$ is 1-to-1 when restricted to $\mathrm{im} X$ (the flat torus) because $f \circ X$ is immersive and by restricting to $u, v \in [0, 2\pi)$, both $f$ and $f \circ X$ become 1-to-1. In other words, if f( X(u,v) ) = f(X(u', v')), then X(u,v) = X(u', v').
Hope this helps!