How to understand (maybe, informally) why the intersection of two quadrics in general position in $\mathbb{CP}^3$ is an elliptic curve?

It is obvious that it is a compact 2-manifold, i.e. a sphere with handles, but how to find its genus? I suppose one can apply Mayer-Vietoris sequence, but the union of two quadrics seems to be hard to realise.

Update: I was told that it is possible to use Riemann-Hurwitz formula. Let us choose two lines $a$ and $l$ and project all points from $a$ to $l$. Than it will be branched covering, with 4 sheets by Bezout's theorem. It shoould be $$0=\chi(\text{intersection})=\chi(l=\mathbb{CP}^1)*\text{sheets}-\text{ramification}=2*4-8,$$ but is there an easy way to see these ramification points?

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Your comments about Mayer-Vietoris seem off kilter a bit (you would need to write the intersection $X=Q\cap Q'$ as a union of two good sets). Although there may be an ad hoc approach, the standard approach is to use the adjunction formula. We compute the canonical bundle in two steps: $$K_Q \cong K_{\Bbb P^3}\otimes\mathscr O_{\Bbb P^3}(2)\big|_Q \cong\mathscr O_Q(-2), \text{ and } K_X \cong K_Q\otimes \mathscr O_Q(2)\big|_X \cong \mathscr O_X.$$ The genus formula for smooth curves tells us that trivial canonical bundle is equivalent to $g=1$.
EDIT: Cool argument you've added with Riemann-Hurwitz. One elementary way to get the $8$ is this: Say the line $\ell$ is given by homogeneous linear equations $L_1=L_2=0$ and the two quadrics are $Q_1=0$ and $Q_2=0$, respectively. You get another quadric surface by taking the equation $$D=\left|\begin{matrix} - & dL_1 & - \\- & dL_2 & - \\- & dQ_1 & - \\- & dQ_2 & - \end{matrix}\right| = 0.$$ The exercise is to show that points of $X\cap D$ correspond to points of the curve $X=Q_1\cap Q_2$ where the projection $\pi_a$ from $X$ to $\ell$ ramifies. By Bezout, since $X$ has degree $4$ and $D$ has degree $2$, we get $8$ such points.