= Solution
At transfer <apoapsis>, <angular momentum conservation> gives
$$
v_a=\frac{r_p}{r_a}v_p=\frac13\sqrt{\frac32}\,v_0=\frac{v_0}{\sqrt6}.
$$
The desired <circular Kepler orbit> has speed $v_c=\sqrt{GM/(3r_0)}=v_0/\sqrt3$. Thus
$$
\Delta v_2=v_0\left(\frac1{\sqrt3}-\frac1{\sqrt6}\right),\qquad
\boxed{f_2=\frac{\Delta v_2}{v_0}=0.1691\simeq16.9\%}
$$
of the original fuel supply is used in the second burn. This is about $21.8\%$ of the fuel remaining after the first burn. The total transfer consumes $f_1+f_2\simeq39.4\%$ of the original supply, leaving about $60.6\%$. The two positive <impulse> magnitudes are added for the fuel budget, even though their directions at opposite ends of the orbit differ.
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