Dear liu,
1) If $A$ is a domain in which every finitely generated ideal is principal, then a module over $A$ is flat iff it is torsion free (Bourbaki, Comm.Alg.,I,§2, 4, Prop.3). Of course a PID has this property, but the ring $\mathcal O(U)$ of holomorphic functions over a connected open subset $U\subset \mathbb C$ also has it, although it is not a PID.[Ah, I have just checked the definition of Prüfer and it seems that these rings are actually Prüfer, so you knew this. I'll leave this class of examples for the benefit of those who, like me, don't know the concept "Prüfer ring".]
2) A simple example of torsion free non flat module $E$ over a ring $A$ is $A=\mathbb C [t^2, t^3] \subset E=\mathbb C [t]$. This corresponds to the normalization of the cusp $S=Spec (\mathbb C[X,Y]/(Y^2-X^3))$ i.e. to the morphism $f: \mathbb A ^1 \to S \subset \mathbb A ^2$ given by $x=t^2, y=t^3$. Non-flatness is due to the fact that the fiber of $f$ at the origin is a double point on $\mathbb A ^1$ (supported at the origin) whereas the other fibers are simple points. This is a tiny particular case of the constancy of Hilbert polynomials in flat families. Of course there are direct proofs by pure algebra, but I hope you like geometry...
The end of Mumford's red book Here is a vast generalization of 2). Consider a locally noetherian integral scheme. If the scheme is not normal, its normalization is never flat over it.This is essentially proved in Matsumura's Commutative Ring Theory, Corollary to Theorem 23.9. (So if you are arithmetically inclined, $\mathbb Z[2i] \subset \mathbb Z[i]$ is a non flat algebra) . It is very easy to find a (slightly different) statement in the literature: the one and a half last lines of Mumford's Red Book !
Edit: I'm happy to report that I just located non-flatness of normalization in our friend Qing Liu's book Algebraic Geometry and Arithmetic Curves : 4.3.1. Example 3.5, page 136.
Best Answer
The best book for such questions in my opinion is the one you're already reading: "Lectures on Modules and Rings" by Lam. Indeed, on page 127 he provides a counter-example to your claim that torsion-free implies flat. Probably you meant the converse, which does hold: Any flat module is torsion-free. This is also on page 127.
Here's Lam's counter-example...Let $R=k[x,y]$ where $k$ is any commutative domain. Then $M=(x,y)$ is torsion-free because there are no relations on $x$ or $y$. However, $M$ is not flat. To see this set $S=R/(x)\cong k[y]$ so that $M\otimes_R S = M\otimes_R R/(x) \cong M/xM \cong (x,y)/(x^2,yx)$. If $M$ is flat over $R$ then $M\otimes_R S$ is flat over $S$ and hence torsion-free. This is a contradiction because $yx=0$ but $y\neq 0$.