In the example in the image you link to, the two signals vary on similar time scales. In that case it's not entirely clear what it means to "mold" one to "fit" the other. The only interpretation I can think of is to make sure that the one doesn't extend beyond the other; if that's what you want, you just need to take minimums and/or maximums, either on one side or, if you want the one signal to be "axis-side" of the other, then on both sides:
$f_3= \max(\min(f_1,|f_2|),-|f_2|)\;.$
But the more usual scenario is that $f_2$ varies significantly less rapidly than $f_1$. In that case, it makes sense to consider $f_2$ as an "envelope" for $f_1$, and to get a version of $f_1$ "shaped" by $f_2$ by multiplying the two. Of course you can do that in your case, too, but it will result in a bit of a mess, frequency-wise, since the two signals have similar frequency ranges and the multiplication will get them all mixed up, whereas if their frequencies are reasonably far apart then it makes sense to still regard the product signal as essentially $f_1$, just with a time-dependent amplitude given by $f_2$.