From Calculus On Manifolds by Spivak
If $f\colon\,\mathbb{R}^n \to \mathbb{R}^n$ is differentiable, then $f^*(h\, dx^1\wedge\dots\wedge dx^n)=(\det f')\cdot(h\circ f)\,dx^1\wedge...\wedge dx^n$
Proof: Since $f^*(h\, dx^1\wedge\dots\wedge dx^n)=(h\circ f)f^*(dx^1\wedge...\wedge dx^n)$ it suffices to show that
$$f^*(dx^1\wedge...\wedge dx^n)=(\det f')\,dx^1\wedge...\wedge dx^n.$$
Let $p \in \mathbb{R}^n$ and let $A=(a_{ij})$ be the matrix of $f'(p)$. (We shall omit "p" in $dx^1_p\wedge\cdots\wedge dx^n_p$ if it causes no confusion).
$$\begin{align}f^*(dx^1\wedge\dots\wedge dx^n)(e_1,...e_n)&=\\ &=(dx^1\wedge\dots\wedge dx^n)(f_*e_1,\dots, f_*e_n)=\\ &=(dx^1\wedge...\wedge dx^n)\left(\sum_{i=1}^{n}a_{i1}e_i,\dots,\sum_{i=1}^{n}a_{in}e_i\right)=\\ &=\det(a_{ij})\,(dx^1\wedge\dots\wedge dx^n)(e_1,...e_n) \end{align}$$
My question is how $f_*e_1=\sum_{i=1}^{n}a_{i1}e_i$?
I know $f_*(v_p)=(Df(p)(v))_{f(p)}=f(p)+Df(p)(v)$.
Let $A\colon V\to W$. Choose a basis $e_1, e_2, \dots, e_n$ of $V$ and $u_1, \dots, u_n$ of $W$. We define the matrix $(a_{ij})$ of $A$ (with respect to the chosen bases) such that $$A(e_j) = \sum_{i=1}^n a_{ij} u_i$$ This is the definition – obviously one needs to show that it makes sense (i.e. numbers $a_{ij}$ exists and are determined uniquely).
In your question $A=f_*=Df_p$ is the derivative, i.e. a linear map $$Df_p\colon T_p\mathbb R^n\to T_{f(p)}\mathbb R^n.$$
Choosing a basis $e_{1,p}, \dots, e_{n, p}$ for $T_p\mathbb R^n$ and $e_{1,f(p)}, \dots, e_{n, f(p)}$ for $T_{f(p)}\mathbb R^n$ (this nomenclature may be a bit confusing) your equation becomes the obvious
$$f_*(e_{j,p}) = \sum_{i=1}^n a_{ij} e_{i, f(p)}.$$
And this is what you should get – you will see this if you add subscripts to $dx^i$ as well.
Edit: I should have said this explicitly, as Ted Shifrin did – the last statement of your question is wrong. Such addition doesn't make sense in general – $f(p)$ is a point of the target manifold and $Df_p(v)$ is a vector tangent to the manifold.