Why is the group of unit upper triangular matrices solvable?

You can prove $U_n$ solvable by induction on $n$. Consider $H_n$, the group of matrices of the form $$ \begin{pmatrix} 1 & & & a_1 \\ & \ddots & & \vdots \\ & & 1 & a_{n-1} \\ & & & 1 \end{pmatrix} $$ It is then easy to see that $H_n \triangleleft U_n$, that $H_n$ is abelian (in fact isomorphic to $k^{n-1}$), and that $U_n/H_n\cong U_{n-1}$.

If $A$ is a $k\times k$ matrix, I call the entries $a_{i,i+\ell}$ of $A$ the $\ell$-th diagonal of $A$. In particular, the $0$-th diagonal of $A$ is its actual diagonal.

You can prove that if $A,B \in U_n(k)$ and $C = [A,B]$ then the first diagonal of $C$ is $0$. More in general, if the first $\ell$ diagonals (except the $0$-th) of $A$ are $0$ and the first $m$ diagonals (except the $0$-th) of $B$ are $0$, then the first $\ell+m+1$ diagonals of $[A,B]$ are $0$.

In order to prove the statement about the quotient, you can observe that $B_n/U_n$ is indeed isomorphic to the group of diagonal matrices.