The problem is wildly underdetermined. Of the infinitely many equations for $J$ of the equation $J^T e=f$ is the (rank-one) matrix
$$
J^T = \frac{f e^T}{\|e\|^2}.
$$
For this, you have that
$$
JJ^T = \frac{e f^Tf e^T}{\|e\|^4} = \|f\|^2 \frac{e e^T}{\|e\|^4}.
$$
Interestingly, this matrix satisfies
$$
(JJ^T)(JJ^T) = \|f\|^4 \frac{e e^T}{\|e\|^4} \frac{e e^T}{\|e\|^4}
= \|f\|^4 \frac{e e^T e e^T}{\|e\|^8}
= \|f\|^4 \frac{e e^T}{\|e\|^6} = \frac{\|f\|^2}{\|e\|^2} JJ^T.
$$
This means that
$$
(JJ^T)^{1/2}
= \frac{\|e\|}{\|f\|} JJ^T.
$$
This allows you to compute the quantity you are interested in.
EDIT: To see how the last line follows, take the square of the left and right hand sides (i.e., multiply each side with itself) to obtain
$$
JJ^T
= \frac{\|e\|^2}{\|f\|^2} (JJ^T) (JJ^T).
$$
This is, up to division by the factor on both sides, the same as before.
So, how to proceed then: We know that $A := JJ^T = c ee^T$ where $c=\|f\|^2/\|e\|^4$. In other words, if you multiply $A$ by $e$, you get a multiple of $c\|e\|^2 e$ from it. Conversely, this means that applying $A^{-1}$ to this direction $e$, you get one over this multiple, i.e., $A^{-1}e=\frac{1}{c\|e\|^2}e$. (This inverse does not strictly exist -- I'm just taking the pseudo-inverse $A^\dagger$ when I write $A^{-1}$). This brings us back to the equation
$$
(JJ^T)^{-1/2}
= \frac{\|f\|}{\|e\|} A^{-1}
$$
and consequently
$$
(JJ^T)^{-1/2}e
= \frac{\|f\|}{\|e\|} A^{-1}e
= \frac{\|f\|}{c\|e\|^2} e
= \frac{\|f\|\|e\|^4}{\|f\|^2\|e\|^2} e
= \frac{\|e\|^2}{\|f\|} e.
$$
With this, you get
$$
J^T(JJ^T)^{-1/2}e
= \frac{1}{\|f\|} fe^Te
= \frac{\|e\|^2}{\|f\|} f.
$$
All of this said, this is just one choice for $J$. Other choices may yield different results.
