Prove that there exists $T in mathcal{L}(V, W)$ such that $operatorname{null}(T) = U$ if and only if $dim (U) ge dim (V) - dim (W)$

First of all your proof does not emphasis where the assumption is used.
To define a linear map $T$, it is sufficient to assign the images of members of a basis. Now we want $text{null}(T)=U$, so first we take a basis $mathcal{B}_0$ of $U$. Then extend $mathcal{B}_0$ to a basis $mathcal{B}$ of $V$. Now to have a linear map $T$ with $text{null}(T)=U$ we have to assign each member of $mathcal{B}_0$ to $0$ and rest of the members of the $mathcal{B}$ have to assign so that ${T(v):vinmathcal{B}smallsetminusmathcal{B}_0}$ is linearly independent in $W$. Now the assumption $dim(U)geqdim(V)-dim(W)$ $implies dim(W)geqdim(V)-dim(U)$ $implies dim(W)geq|mathcal{B}smallsetminusmathcal{B}_0|$. Thus this assures there are $|mathcal{B}smallsetminusmathcal{B}_0|$ many linearly independent vectors in $W$. Which guarantees the existence of required $T$.

Let's first consider this definition: $$T(a_1u_1 + ldots + a_mu_m + b_1v_1 + ldots + b_nv_n) = b_1w_1 + ldots + b_nw_n.$$ This definition only really makes sense because $(u_1, ldots, u_m, v_1, ldots v_n)$ is a basis for $V$. So, any vector $x in V$ can be expressed uniquely in the form $$x = a_1u_1 + ldots + a_mu_m + b_1v_1 + ldots + b_nv_n.$$ Now, if $x in U$, then $x$ must uniquely take the form $$x = a_1u_1 + ldots + a_mu_m = a_1u_1 + ldots + a_mu_m + 0v_1 + ldots + 0v_n,$$ since $(u_1, ldots, u_m)$ is a basis for $U$. So, according to our definition of $T$, for $x in U$, we have begin{align*} T(x) &= T(a_1u_1 + ldots + a_mu_m) = T(a_1u_1 + ldots + a_mu_m + 0v_1 + ldots + 0v_n) \ &= 0w_1 + ldots + 0w_n = 0. end{align*} So, $U subseteq operatorname{Null} T$.

Conversely, suppose $x in operatorname{Null} T$. We still know $x$ is in the form $$x = a_1u_1 + ldots + a_mu_m + b_1v_1 + ldots + b_nv_n,$$ but this time we know that $$0 = T(x) = b_1w_1 + ldots + b_nw_n.$$ Therefore, $$x = a_1u_1 + ldots + a_mu_m + 0 in U,$$ completing the proof that $U = operatorname{Null} T$.

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Why define it with $b_1, ldots, b_n$ instead of $c_1, ldots, c_n$? Well, remember that $b_1, ldots, b_n$ are not constants, they are placeholder variables. $T$ is defined by expansion with respect to the basis $(u_1, ldots, u_m, v_1, ldots, v_n)$, and the way that the author has chosen to denote such an expansion is by $$x = a_1u_1 + ldots + a_mu_m + b_1v_1 + ldots + b_nv_n.$$ Thus, $b_1, ldots, b_n$ are defined implicitly as (linear) functions of $x$, taking the vector $x$, and returning the coordinate of the corresponding basis vector $v_i$.

To simply replace them with $c_1, ldots, c_n$, without defining them somehow, would mean the transformation is ill-defined. What are $c_1, ldots, c_n$ in this context? How does the value of $x$ change them?

You could replace $b_1, ldots, b_n$ with certain functions of $b_1, ldots, b_n$ to obtain an equally valid construction $T'$ such that $operatorname{Null} T' = U$. For example, the following $T'$ will also work: $$T'(a_1u_1 + ldots + a_mu_m + b_1v_1 + ldots + b_nv_n) = b_nw_1 + ldots + b_1w_n.$$ Such functions are usually not unique!

Another way to write it:

Define $T : V to W$ by $T(u_i) = 0_W$ for $i in {1,dots,m}$, $T(v_j) = w_j$ for $j in {1,dots,n}$ and extend it by linearity.

Note that the hypotheses implies that $$p = dim(W) geq dim(V) - dim(U) = (m+n)-m = n,$$ so choosing $w_1,dots,w_n$ of $w_1,dots,w_p$ makes sense.

Also, note that this satisfies your definition, for if $a_1,dots,a_m,b_1,dots,b_n in mathbf F$, then begin{align} T(a_1u_1 + & cdots + a_mu_m + b_1v_1 + cdots + b_nv_n) \ &= a_1T(u_1) + cdots + a_mT(u_m) + b_1T(v_1) + cdots + b_nT(v_n) \ &= b_1w_1 + cdots + b_nw_n. end{align}

Now, in one hand, it is easy to see that $U subseteq operatorname{null}(T)$ since every $u in U$ can be written as a linear combination of $u_1,dots,u_m$. On the other hand, let $v in V$ and write it as $$v = c_1u_1 + cdots + c_mu_m + d_1v_1 + cdots + d_nv_n$$ for some $c_1,dots,c_m,d_1,dots,d_n in mathbf F$. If $v in operatorname{null}(T)$, then $$0_W = T(v) = d_1w_1 + cdots + d_nw_n$$ and since $w_1,dots,w_n$ are linearly independent, $d_1 = cdots = d_n = 0$. So $$v = c_1u_1 + cdots + c_mu_m in U.$$