Prove $\exists (\varphi_1, \ldots, \varphi_m) \in \mathcal{L}(V, \mathbf{F}) : T(v) = \varphi_1(v)w_1 + \cdots + \varphi_m(v)w_m$, $\forall v \in V$ 
Suppose $T \in \mathcal{L}(V, W)$, and $(w_1, \ldots, w_m)$ is a basis of $\operatorname{range}(T)$. Prove that there exists $(\varphi_1, \ldots, \varphi_m) \in \mathcal{L}(V, \mathbf{F})$ such that $$ T(v) = \varphi_1(v)w_1 + \cdots + \varphi_m(v)w_m $$ for every $v \in V$.

$V$ and $W$ are vector spaces over a field $F$.
My initial attempt to this question is as follows:
$(w_1, \ldots, w_m)$ is a basis for $\operatorname{range}(T)$. Therefore we can write $T(v)$ as $$ T(v) = \alpha_1 w_1 + \cdots + \alpha_m w_m $$ where $\alpha_1, \ldots, \alpha_m \in \mathbf{F}$.
Define $\varphi_i(v) = \alpha_i$ for $i=1, \ldots,m$.
At this point I became stuck. How am I suppose to show that for every choice of $v$, my definition for $\varphi_i$ is going to give the correct value or $\alpha_i$?
After looking at the answer, they just continue on from where I became stuck to show all $\varphi_i$ are linear maps. But, this just shows that $\varphi_i$ is a linear map, not that it exists or satisfies the given condition?

Here's a slightly different approach:
Given $v\in V$, let $\Phi(v)\in F^m$ be the unique $m$-tuple such that
$$
T(v)=(\Phi(v))_1w_1+\dots+(\Phi(v))_mw_m
$$
where we write $x_i$ for the $i$th entry of the tuple $x$. This definition makes sense because $w_1,\dots,w_m$ form a basis.
Let $\varphi_i(v):=(\Phi(v))_i$.
You can either check that $\varphi_i$ is linear by hand, or argue that $\Phi$ is linear and that $\varphi_i=\pi_i\circ \Phi$, where $\pi_i:F^m\to F$ is the $i$th projection, which is also linear.
