In Introduction to Smooth Manifolds by John M. Lee p.360, he defines $\Omega^*(M) =\displaystyle\bigoplus_{k=0}^n\Omega^k(M)$, the differential forms on a manifold $M$, as being an associative, anticommutative graded algebra.
My understanding is that this an algebra over the ring of smooth functions $C^\infty(M)$, and not an algebra over the field $\mathbb{R}$.
In graded algebras over a field, each $\Omega^k(M)$ is a vector space, and as such has a basis.
But in our case, all we can say is that $\Omega^k(M)$ is a module over $C^\infty(M)$. I wish I could say, for instance, that I have a basis for $\Omega^1(M)$ given by $dx^1,\cdots, dx^n$. However, since this is not a vector space I am not sure how to make this idea formal, because talking about dimension is not as straight-forward in modules. Is it right to say that $\Omega^1(M)$ is a finitely-generated module over $C^\infty(M)$ (with perhaps the "basis" being $dx^1,\cdots, dx^n$)?
Ok, let's take the algebra $\Omega^1(M)$ of differential 1-forms (just to simplify the discussion, it's the same for every other $\Omega^k(M)$). First of all, nitpicking: $\Omega^*(M)$ does have a structure making it an algebra over $\mathbb R$, since $\mathbb R$ is a subring of $C^\infty(M)$ (consider the constant functions $M\rightarrow\mathbb R$). In general, if you have a $S$-algebra structure on a module $X$ and a subring $R\hookrightarrow S$, then there is also a $S$-algebra structure on $M$.
Now, as said in the comments, the differentials $dx^1,\cdots,dx^n$ are not globally defined differential 1-forms, but are locally defined. So, if $\{(U_\alpha,x_\alpha)\}_\alpha$ is you atlas for $X$, then each differential 1-form $\omega$ restrict to a differential 1-form $\omega_\alpha=\omega|_{U_\alpha}\in\Omega^1(U_\alpha)$ (for any $\alpha$). Now, for every $U_\alpha$ you have a diffeomorphism $$T^*U_\alpha\rightarrow U_\alpha\times\mathbb R^n$$ (if $x_\alpha$ are coordinates $\phi:U_\alpha\rightarrow\mathbb R^n$) and this diffeomorphism preserves fibers (that is, if $\delta\in T^*_pU_\alpha\rightarrow \phi(\delta)\in\{p\}\times\mathbb R^n$). The diffeomorphism $\phi$ is also fiberwise linear, that is, it induces a linear map $T^*_pU_\alpha\rightarrow\{p\}\times\mathbb R^n$, for all $p\in U_\alpha$. So, composition with $\phi$ induces a $C^\infty(U_\alpha)$-linear isomorphism $$\Omega^1(U_\alpha)\xrightarrow\simeq\mbox{Sections}(U_\alpha\times\mathbb R^n\rightarrow U_\alpha)\simeq C^\infty(U_\alpha,\mathbb R^n)\simeq C^\infty(U_\alpha,\mathbb R)^n$$ so $\Omega^1(U_\alpha)$ is a free $C^\infty(U_\alpha)$-module. the second isomorphism is, i think, clear: A map $\sigma:U_\alpha\rightarrow U_\alpha\times\mathbb R^n$ mapping $p$ into $\{p\}\times\mathbb R^n$ is determined by the composition $$U_\alpha\xrightarrow\sigma U_\alpha\times\mathbb R^n\xrightarrow{\pi}\mathbb R^n$$ and every map $U_\alpha\rightarrow\mathbb R^n$ is determined by the $n$-tuple of its components (which is the third isomorphism). In the isomorphism $\Omega^1(U_\alpha)\simeq C^\infty(U_\alpha)^n$, the differential $dx_\alpha^j$ corresponds to the tuple $$(0,0,\cdots,\underbrace{1}_{j\text{th component}},0,\cdots,0)$$where $1:U_\alpha\rightarrow \mathbb R$ is the function constant $=1$.
So there you have it: The algebra $\Omega^1(M)$ may not be a free $C^\infty(M)$-module, but there is a cover $\{U_\alpha\}_\alpha$ such that $\Omega^1(U_\alpha)$ is a free $C^\infty(U_\alpha)$-module. And by the restriction map $\Omega^1(M)\rightarrow\Omega^1(U_\alpha)$ (given by $\omega\mapsto\omega|_{U_\alpha}$, every differential 1-form $\omega\in\Omega^1(M)$ locally a $C^\infty(U_\alpha)$-linear combination of the differentials $dx_\alpha^1,\cdots,dx_\alpha^n$, for every $\alpha$