Extending linear maps from subspaces to the entire space This is from Axler's Linear Algebra Done Right, 3.A.11

Suppose $V$ is finite-dimensional. Prove that every linear map on a
subspace of $V$ can be extended to a linear map on $V$.  In other words,
show that if $U$ is a subspace of $V$ and $S \in L( U, W$), then there exists
$T \in L (V,W)$ such that $T(u) = S(u) $ for all $u \in U$.

I tried to prove this by creating another transformation that took all vectors into U and then mapped all vectors in $U$ into $S$($U$). (So a double transformation I guess.)
The answer book had  a different solution which I didn't understand:

I don't see how this is linear, because consider $u + v$, where $u \in U$, $S$($u$) $\ne 0 $ and $v \in V$ and $u + v \in V \notin U $. (since $U \subset V$).
Then $T$($u+v$) $=0$, but   $T$($u+v$) $=0$, but  $T$($u$) $+  T$($v$) $=$ $S$($u$) $\ne 0 $.
Which means that T is not linear.
 A: This is really a matter of applying the theorem you asked about in your previous question. Your previous theorem said (I modify the notation slightly)

Let $V$, W be vector spaces over a field $F$, and let $\{\xi_1, \dots, \xi_n\}$ be a basis of $V$. Also, choose $n$ vectors $w_1, \dots, w_n \in W$. Then, there is a unique linear transformation $T:V \to W$ such that for every $i \in \{1, \dots, n\}$, we have $T(\xi_i)= w_i $

So, now you started with a basis $\{u_1, \dots, u_m\}$ of $U$, and then you extend it to a basis $\{u_1, \dots, u_m, v_{m+1}, v_n\}$ of $V$. Now, to make it super explicit how to directly apply the theorem stated above, let's rename things: define
\begin{equation}
\xi_i =
\begin{cases}
u_i& \text{if $1 \leq i \leq m$} \\
v_i & \text{if $m+1 \leq i \leq n$}
\end{cases}
\end{equation}
and define
\begin{equation}
w_i =
\begin{cases}
S(u_i)& \text{if $1 \leq i \leq m$} \\
0 & \text{if $m+1 \leq i \leq n$}
\end{cases}
\end{equation}
Then, by the theorem, there exists a (unique) linear map $T: V \to W$, such that for all $i \in \{1, \dots, n\}$, $T(\xi_i) = w_i$. 
Now, to show $T$ restricted to $U$ equals $S$, i.e for all $x \in U$, $T(x) = S(x)$, notice that BY DEFINITION, for all $i \in \{1, \dots, m\}$ we have that $T(u_i) = S(u_i)$. Since $T$ and $S$ agree on the basis $\{u_1, \dots u_m\}$ of $U$, they agree everywhere on $U$ (this is the uniqueness argument I presented in the previous answer).
