How to prove that a pullback map is linear The following question was left as an exercise in my assignment of Manifolds and I am not able to prove this.
Question: Define the map $T^{*} : L^{k}(W) \to L^{k} (V)$  , where $\alpha \in L^{k}(W)$ defined by $T^{*} (\alpha) ( v_1,...,v_k) = \alpha( T(v_1),...,T(v_k))$. $T^{*} (\alpha)$ is called pullback map of $\alpha$ by $T^{*}$.
Here $L^k (V)= V^{*} \oplus ...\oplus V^{*}$.
$T^{*} (\alpha) (v_1+ w_1,...,v_k +w_k)= \alpha ( T(v_1+w_1) ,...,T(v_k+w_k)) = \alpha( T(v_1) +T(w_1), ..., T(v_k) +T(w_k))$.

But I am not able to prove the RHS equal to $\alpha(T(v_1) ,..., T(v_k) )$ + $\alpha( T(w_1,...,T(w_k))$.

Can you please help me with this?
 A: Suffices to explore $k=2$. This is the only way I can make sense of this picture.
I am assuming $\alpha \in W^*\oplus W^*$ means $\alpha : W\oplus W \to \mathbb K$ is bilinear. This means that $T^*\alpha$ is bilinear map, not linear. So, in general, $\alpha \in L^k(W)$ is a $k$-linear map.
Assume $T:V\to W$ is linear. Then
$$
\begin{align*}
 (T^*\alpha)(\lambda u_1+\mu v_1,u_2) &= \alpha (T(\lambda u_1+\mu v_1), Tu_2) \\
&= \alpha (\lambda Tu_1 + \mu Tv_1, Tu_2) \\
&= \alpha (\lambda Tu_1,Tu_2) + \alpha (\mu Tv_1,Tu_2) \\
&= \lambda\alpha (Tu_1,Tu_2) + \mu\alpha (Tv_1,Tv_2) \\
&= \lambda(T^*\alpha)(u_1,u_2) + \mu(T^*\alpha)(v_1,u_2)
\end{align*}
 $$
and similarly for linearity in second component.
And for linearity of $T^*$ one has
$$
\begin{align*}
T^*(\lambda\alpha + \mu\beta)\mathbf v &= (\lambda\alpha+\mu\beta)(Tv_1,Tv_2) \\& = \lambda\alpha(Tv_1,Tv_2) + \mu\beta (Tv_1,Tv_2)\\ &= \lambda(T^*\alpha)\mathbf v + \mu(T^*\beta)\mathbf v \\
&= (\lambda T^*\alpha + \mu T^*\beta)\mathbf v
\end{align*}
 $$
for every $(v_1,v_2) =:\mathbf v \in V\oplus V$.
