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 $y\neq 0$.