TheFirst, let's clarify what the statement means: The Steenrod algebra is a graded algebra, and $H^*(X)$ for any spectrum $X$ is a graded module over it. This means that the action $\mathcal{A}_2\otimes H^*(X)\to H^*(X)$ takes $\mathcal{A}_2^n \otimes H^m(X) \to H^{n+m}(X)$. So yes, it does not make sense to consider any individual $H^m(X)$ as module over the Steenrod algebra.
What does it mean to be a free module, in this graded world? You can write down the "free module over $\mathcal{A}_2$ on a generator in degree $n$", this is just the $\mathcal{A}_2$-module which is given in degree $m$ by $\mathcal{A}_2^{m-n}$. Let's write that as $\mathcal{A}_2(n)$. Now we call a graded $\mathcal{A}_2$-module $M$ free if it is isomorphic to a direct sum $\bigoplus_i \mathcal{A}_2(n_i)$.
This is for example the case for $H^*(MO)$ (this is due to Thom originally, I believe). It is not quite the case for $H^*(MO(r))$, but you get that it "looks like a free module in degrees $\leq 2r$". What does this mean? It means you find a map $\bigoplus_i \mathcal{A}_2(n_i) \to H^*(MO(r))$ of graded modules, which is an isomorphism in degrees $\leq 2r$.
Now as for how this is justified (without having double-checked that this is the way Stong does it): In principle, all of this can be verified explicitly, by using the Thom isomorphism, a description of the cohomology of $BO(r)$ with Steenrod action, and the fact that the Thom isomorphism twists the Steenrod action by Stiefel-Whitney classes. But the quote at the end of your question leads me to believe that Stong deduces the unstable statement (for $MO(r)$) from the stable one (for $MO$). The stable case is originally done by Thom through explicit computations, but there is a more conceptual argument based on the Milnor-Moore theorem. This is explained nicely in the Blog of Akhil Mathew, for example. And the stable statement implies the quoted statement for $MO(r)$, since $H^*(MO)$ in the range of degrees $[0,r]$ is isomorphic to $H^*(MO(r))$ in the range of degrees $[r,2r]$.