= Solution
Define the <pointwise representation of one-forms by vector-field functionals> by
$$
\boxed{\theta(\omega)(X)=\omega(X).}
$$
It is $\mathbb R$-linear, its value is a smooth function, and $X_p=0$ implies $\theta(\omega)(X)(p)=0$. Moreover
$$
\theta(g\omega)=g\,\theta(\omega),
$$
so it is a map of $C^\infty(M)$-modules. This is the properly typed scalar-multiplication property: the same symbol for a form and its associated functional is understood through $\theta$.
To construct its inverse, let $\alpha\in W$ and $v\in T_pM$. Choose a global smooth <vector field> $X$ with $X_p=v$; a local coordinate field multiplied by a <smooth bump function> supplies such an extension. Set
$$
\omega_p(v)=\alpha(X)(p).
$$
This is well-defined: if $X_p=Y_p$, then $X-Y$ vanishes at $p$, so $\alpha(X-Y)(p)=0$. Real linearity of $\alpha$ makes $\omega_p$ a <covector>. In fact the pointwise vanishing assumption also forces
$$
\alpha(fX)(p)=f(p)\alpha(X)(p),
$$
because $(f-f(p))X$ vanishes at $p$. Thus $\alpha(fX)=f\alpha(X)$ for every smooth $f$, even though only real linearity was initially imposed.
It remains to prove smoothness, rather than assume continuity of an abstract linear map. Near a fixed point, choose global fields $E_1,\ldots,E_n$ equal to the coordinate basis on a smaller neighborhood, using a bump function equal to one there. On that neighborhood,
$$
\omega=\sum_i\alpha(E_i)\,dx^i.
$$
All coefficients are smooth by the definition of $W$, so $\omega$ is a smooth <differential one-form>. The construction holds near every point, hence gives $\omega\in\Omega^1(M)$ with $\alpha(X)=\omega(X)$ globally.
Every tangent vector can be realized by a global field, so $\theta(\omega)=0$ forces $\omega=0$. The inverse just constructed gives surjectivity. \b[The evaluation map is a natural $C^\infty(M)$-module isomorphism $\Omega^1(M)\cong W$.] No auxiliary metric or connection was chosen.
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