In a typical differential geometry course,when we study integration over manifolds,then we define first something called the pullback of a differential-$k$ form.Suppose $f:U\subset \mathbb R^n\to V\subset \mathbb R^m$ is a $C^\infty(U)$ function and $\omega$ is a differential $k$ form on $V$,i.e. $\omega\in \Omega^k(V)$.Then we define the pullback of $\omega$ via $f$ to be $f^*:\Omega^k(V)\to \Omega^k(U)$ defined by,
$f^*\omega(p)(v_1,v_2,...,v_k)=\omega_{f(p)}(Df_p(v_1),Df_p(v_2),...,Df_p(v_k))$.
Is this definition made in this way because we want the change of variables formula to hold?This is how I think of it.But I want to get a clearer intuition.
Let $\Sigma$ and ${M}$ be smooth manifolds with $\mathrm{dim}(\Sigma)\leqslant\mathrm{dim}({M})$, further let ${f}:\Sigma\longrightarrow{M}$ be a diffeomorphism. Then ${f}^{\star}:\Omega({M})\longrightarrow\Omega(\Sigma)$ is the pullback of differential forms under that diffeomorphism. Now consider a differential form $\omega\in\Omega^{2}({M})$, as well as the coordinate charts $\xi$ on $\Sigma$ and ${x}$ on ${M}$. Then the pullback is defined as \begin{align*} ({f}^{\star}\omega)_{\alpha\beta}&=({f}^{\star}\omega)\bigg\langle\frac{\partial}{\partial\xi^{\alpha}},\frac{\partial}{\partial\xi^{\beta}}\bigg\rangle\\[0.5em] &=\omega({f}(-))\Bigg\langle\bigg({f}_{\star}\frac{\partial}{\partial\xi^{\alpha}}\bigg)_{{f}(-)},\bigg({f}_{\star}\frac{\partial}{\partial\xi^{\beta}}\bigg)_{{f}(-)}\Bigg\rangle\\[0.5em] &=\omega({f}(-))\Bigg\langle\bigg(\frac{\partial({x}^{\mu}\circ{f})}{\partial\xi^{\alpha}}\bigg\vert_{{f}^{-{1}}(-)}{\,}\frac{\partial}{\partial{x}^{\mu}}\bigg)_{{f}(-)},\bigg(\frac{\partial({x}^{\nu}\circ{f})}{\partial\xi^{\beta}}\bigg\vert_{{f}^{-{1}}(-)}{\,}\frac{\partial}{\partial{x}^{\nu}}\bigg)_{{f}(-)}\Bigg\rangle\\[0.5em] &=\omega({f}(-))\Bigg\langle\frac{\partial({x}^{\mu}\circ{f})}{\partial\xi^{\alpha}}\bigg(\frac{\partial}{\partial{x}^{\mu}}\bigg)_{{f}(-)},\frac{\partial({x}^{\nu}\circ{f})}{\partial\xi^{\beta}}\bigg(\frac{\partial}{\partial{x}^{\nu}}\bigg)_{{f}(-)}\Bigg\rangle\\[0.5em] &=\frac{\partial({x}^{\mu}\circ{f})}{\partial\xi^{\alpha}}\frac{\partial({x}^{\nu}\circ{f})}{\partial\xi^{\beta}}{\,}\omega({f}(-))\Bigg\langle\bigg(\frac{\partial}{\partial{x}^{\mu}}\bigg)_{{f}(-)},\bigg(\frac{\partial}{\partial{x}^{\nu}}\bigg)_{{f}(-)}\Bigg\rangle\\[0.5em] &=\frac{\partial({x}^{\mu}\circ{f})}{\partial\xi^{\alpha}}\frac{\partial({x}^{\nu}\circ{f})}{\partial\xi^{\beta}}{\,}\omega_{\mu\nu}\circ{f}{\,}{.} \end{align*} One can now obeserve that this expression of the pulback is a generalization of the coordinate transformation law for the component functions of a tensor field under a change of coordinate charts. The pullback is important in order to understand the geometry of embeddings $\Sigma\longrightarrow{M}$, since the pullback of the metric ${g}\in\Gamma^{\infty}({M},{T}^{\star}{M}\otimes{T}^{\star}{M})$ on ${M}$ yields the induced metric ${f}^{\star}{g}\in\Gamma^{\infty}({\Sigma},{T}^{\star}{\Sigma}\otimes{T}^{\star}{\Sigma})$ on ${f}(\Sigma)$ which can be used to calculate the volume \begin{align*} \mathrm{vol}({f}(\Sigma))=\int_{\Sigma}\sqrt{\mathrm{det}(({f}^{\star}{g})_{\alpha\beta})}{\,}\mathrm{d}\xi{\,}{.} \end{align*}