If you already know that you can compete a basis of a subspace to
a basis for the whole space then you are practically done.
Hint: Note that $\alpha$ is a basis for $V$ (this should
give you $W_{1}+W_{2}=V$, why ?) and that the $w_{i}$ are linearly
independent of the $u_{i}$ (this should show that $W_{1}\cap W_{2}=\{0\}$,
why ?)
Note: The way I see it, there is no use for $\beta$ or of the $v_{i}$
in the proof
Let us construct a function $f:W_1 \to W_2$ as follows.
Given any $w \in W$, it has a unique representation as
$$w = w_1 + w_2,$$
where $w_1 \in W_1$ and $w_2 \in W_2$.
We define $f(w_1) = w_2$. It is clear that this $f$ does actually satisfy the condition you want. That is, $W = \{w_1 + f(w_1) \mid w_1 \in W_1\}$.
However, what is still left is to show that $f$ is indeed a well-defined linear function.
First, we show given any $w_1 \in W,$ there do exist vectors $w \in W$ and $w_2 \in W_2$ such that $w = w_1 + w_2$, that is, $f(w_1)$ has some value.
This is easy as $w_1 \in W_1 \subset V = W \oplus W_2.$ Thus, $w_1 = w + w_2$ for some $w \in W$ and some $w_2 \in W_2$. Rearranging the equation and using the fact that $-w_2 \in W_2$ gives us the desired result.
Secondly, we show that there is no ambiguity with this choice of $f(w_1).$
Suppose that $w_1 + w_2 = w \in W \ni \widehat{w} = w_1 + \widehat{w_2}$ for some $w_2, \widehat{w_2} \in W_2.$ We want to show that $w_2 = \widehat{w_2}.$
Note that $w - \widehat{w} = w_2 - \widehat{w_2}.$
The LHS is an element of $W$ and the RHS of $W_2.$ Thus, we get that $w_2 - \widehat{w_2} \in W \cap W_2 = \{0\}$, that is, $w_2 = \widehat{w_2}$, as desired.
Thirdly, we show that $f$ is indeed a linear map.
Suppose $x, y \in W_1$ and $\alpha \in \mathbb{F}$, the field over which $V$ is a vector space.
By construction, we have that $\alpha x + f(\alpha x) = X \in W$ and $y + f(y) = Y \in W$.
Thus, $\underbrace{\alpha x + y}_{\in W_1} + \underbrace{f(\alpha x) + f(y)}_{\in W_2} = \underbrace{X+Y}_{\in W}.$
By our definition, we get that $f(\alpha x + y) = f(\alpha x) + f(y)$ and hence, $f$ is a linear map.
Best Answer
Since $W_1\subseteq V$ and $W_2\subseteq V$, $W_1+W_2\subseteq V$.
Since $W_1\cap W_2=\{0\}$, the sum above is a directe sum. So $\dim(W_1)+\dim(W_2)\le\dim V$.
$V=W+W_1$ implies $\dim(V)\le \dim(W)+\dim(W_1)$.
$V=W+W_2$ implies $\dim(V)\le \dim(W)+\dim(W_2)$.
Adding all inequalities above and cancelling out, we get $$\dim V\le2\dim(W),$$ which is, in fact, an equality as given. Hence all inequalities above must be equalities.
Now you can try applying the second statement you found in your book.