The material on this page is taken mostly from papers (written in French) by Mestre [1] and Fermigier [2]. The 'for dummies' part comes in that I've elided most of the justification and the algebraic-geometry work, and given Maple implementations of the easier algorithms.
The easiest construction is given as 'Method 2' in Mestre's first paper. Fix some field F, and let q(x) be a monic degree-6 polynomial with roots in F. Using another indeterminant t, define p(x) = q(x+t) q(x-t), so p is a monic degree-12 polynomial in x with coefficients in F[t].
Now let Q(x) = x6 + a5x5 + a4x4 + a3x3 + a2x2 + a1x + a0, a general degree-6 monic polynomial, and consider R = Q2 - p as a polynomial in x with coefficients in F[t]. Note that the coefficient of x6+j depends only on ak for k>=j, so if we successively equate the coefficients of x11, ..., x6 in R to zero, we can solve in succession for a5 ... a0 and, at each stage, have only a linear equation to solve. Effectively, we're computing a symbolic square-root of p, and letting R be the remainder term; the result is the same as computing the polynomial part of the asymptotic expansion of p1/2 (using, for example, the Maple command asympt), but more explicit. You will find that R has a factor t2; remove this, and you discover that the coefficient of x5 in R is dependent only on the roots of q.
Pick the roots of q carefully so that this coefficient is zero; working over Z, Mestre finds that roots -17, -16, 10, 11, 14, 17 happen to work.
Now consider the curve y2 = R(x). This is a quartic curve (because we have killed the coefficient of x5 in R) with coefficients in F[t]; pick some value of t such that the roots of p are all distinct, and specialise to get a curve over F. Since R = Q2-p, we know that, if we take x as a root of p, we'll have R(x) a square - and so [x, sqrt(R(x))] is a point on the curve. So we have y2 equalling a quartic with 12 distinct rational points - but, even if we had only one rational point, we could pick that as the origin and convert the quartic to an elliptic curve in Weierstrass form. So we get up to 12 elliptic curves (by picking 12 different points as the origin), each with 11 distinct rational points. In fact, you can show using algebraic geometry that the rational points considered with co-ordinates in F[t] are independent, so we get an elliptic curve with 11 independent rational points, and so of rank at least 11, for each choice of t.
We haven't proved that the curves are distinct, but if you compute the curve for two t values you find different ones; similarly, if you don't want to use the algebraic-geometry proof that the rational points are independent, you can fix t and show that the particular points on the particular curve you get are independent.
There's nothing preventing some choice of t from giving more rational points than the construction requires, so you can run over many choices of t, look for points of small height on the quartic which are not the ones given by the construction (you have to search on the quartic, since moving to the Weierstrass model of the curve increases the heights of the points drastically), and check their independence using the height-pairing matrix or the Z2-linear-algebra technique from Theorem 3 of Fermigier's later paper.
Here's a Maple program for constructing R given q.
[1]: Mestre, Courbes elliptiques de rang >=11 sur Q(t), C. R. Acad. Sc. Paris I 313 (1991) pp 139-142.
[2]: Fermigier has published various pages on this subject; see here