If $T:E \to F$ is a linear transformation, $f : F\times\cdots\times F \rightarrow \mathbb{R}$ is an alternating, multilinear $n$-form and $\overline{T}:A_n(F)\rightarrow A_n(E)$ is a function that assigns an alternating, multilinear $n$-form in $E$ to every alternating, multilinear $n$-form in $F$, and is defined by:
$$\overline{T}(f)(v_1,\cdots , v_n) = f(Tv_1, \cdots, Tv_n).$$
How do I know that $\overline{T}$ is linear?
If $S : E \to F$ is also a linear transformation, is it true that $\overline{T+S} = \overline{T}+\overline{S}$?
I really don't see how the second affirmation is true.
First of all,
\begin{align*} \overline{T}(f+g)(v_1, \dots, v_n) &= (f+g)(Tv_1, \dots, Tv_n)\\ &= f(Tv_1, \dots, Tv_n) + g(Tv_1, \dots, Tv_n)\\ &= \overline{T}(f)(v_1, \dots, v_n) + \overline{T}(g)(v_1, \dots, v_n) \end{align*}
so $\overline{T}(f+g) = \overline{T}(f) + \overline{T}(g)$. That is, $\overline{T}$ is linear.
It is not true that $\overline{T+S} = \overline{T} + \overline{S}$. Let $f$ be a non-zero $n$-form on $E$ where $n$ is even, and let $T = \operatorname{id}_E : E \to E$ and $S = - T$. As $T + S$ is the zero map, we have
$$\overline{T+S}(f)(v_1, \dots, v_n) = f((T+S)v_1, \dots, (T+S)v_n) = f(0, \dots, 0) = 0.$$
On the other hand,
\begin{align*} (\overline{T}+\overline{S})(f)(v_1, \dots, v_n) &= \overline{T}(f)(v_1, \dots, v_n) + \overline{S}(f)(v_1, \dots, v_n)\\ &= f(Tv_1, \dots, Tv_n) + f(Sv_1, \dots, Sv_n)\\ &= f(v_1, \dots, v_n) + f(-v_1, \dots, -v_n)\\ &= f(v_1, \dots, v_n) + (-1)^nf(v_1, \dots, v_n)\\ &= f(v_1, \dots, v_n) + f(v_1, \dots, v_n)\\ &= 2f(v_1, \dots, v_n)\\ &\neq 0 \end{align*}
for some choice of $v_1, \dots, v_n$ (as $f$ is non-zero).