My formula of the pullback is given by
If $f:M\longrightarrow N$, then \begin{align*} f^*:\Omega ^k(N)&\longrightarrow \Omega ^k(M)\\ \omega &\longmapsto f^*\omega \end{align*}
where $$(f^*\omega )_p(X_1,...,X_k)=\omega _{f(p)}(\mathrm d f(X_1),...,\mathrm d f(X_k)).$$
I recall that $$\Omega ^k(U)=\{\theta\in Tens_k(U)\mid \forall \sigma \in \mathfrak S_k, {}^\sigma \theta=sgn(\sigma)\theta\}=\{k-differential\ form\ on\ U\}$$ where $sgn:\mathfrak S_k\longrightarrow \{-1,1\}$ is the signature.
I don't understand how to use it... For example, the transformation to polar coordinate to cartesian coordinate, i.e. $$f(r,\theta)=(x,y)=(r\cos\theta, r\sin \theta),$$ how can I compute $f^*\mathrm d x$ using this formula ? I would have $$(f^*\mathrm d x)_{(r,\theta)}\left(\frac{\partial }{\partial r},\frac{\partial }{\partial \theta}\right)=\mathrm d x_{f(r,\theta)}\left(\mathrm d f\left(\frac{\partial }{\partial r}\right), \mathrm d f\left(\frac{\partial }{\partial \theta}\right)\right) $$
By definition, $$\mathrm d f\left(X\right)=X(f)$$ where $X$ is a derivation, and thus I get $$\mathrm d x_{f(r,\theta)}\left(\mathrm d f\left(\frac{\partial }{\partial r}\right), \mathrm d f\left(\frac{\partial }{\partial \theta}\right)\right) =\mathrm d x_{(x,y)}\left(\frac{\partial f}{\partial r},\frac{\partial f}{\partial \theta}\right)=\mathrm d x_{(x,y)}\Big((\cos\theta,\sin \theta),(-r\sin\theta, r\cos\theta)\Big)=???$$
What's next ? (For your information, I know that $$f^*\mathrm d x=\cos\theta\mathrm d r-r\sin \theta\mathrm d \theta$$ but I would like to arrive at this result by using the pullback.
$df_{r,\theta} =\pmatrix{cos\theta & -rsin\theta\cr sin\theta & rcos\theta}$.
Let $(u,v)$ be vector tangent to $(r,\theta)$, $df_{r,\theta}(u,v)=\pmatrix{cos\theta & -rsin\theta\cr sin\theta & rcos\theta}\pmatrix{u\cr v} = (cos\theta u-rsin\theta v,sin\theta u+ rcos\theta v)$.
Thus $f{^*dx}_{r,\theta}(u,v)=dx(df_{r,\theta}(u,v))=dx(cos\theta u-rsin\theta v,sin\theta u+ rcos\theta v) =cos\theta u-rsin\theta v=cos\theta dr(u,v)-rsin\theta d\theta(u,v)$