Y is a subspace of a finite-dimensional linear space X. Let $y_1,...y_j$ be a basis for Y, j=dimY. Then this set can be completed to form a basis for X by adjoining $x_{j+1}, ...,x_n$, n=dimX. Since $y_1,...,x_n$ form a basis for X, every x in X can be expressed as x=$\sum a_i y_i + \sum b_k x_k$. It follows that [x]=$\sum b_k [x_k]$, where [$\cdot$] denotes cosets.
I have trouble with the last sentence. Here is my work:
$\sum b_k [x_k]=\sum b_k (x_k +Y)=\sum_{j+1}^{n} b_k x_k + Y$.
However, since {$x_{j+1},... x_n$} does not span X, i.e. x$\neq \sum b_k x_k$, I cannot understand how [x]=$\sum b_k [x_k]$.
If $y\in Y$ then $[x+y]=[x]$.
Proof. By definition, $$[u]=\{u+v\mid v\in Y\}\ .$$ So if $z\in[x+y]$ then $$\eqalign{ &z=(x+y)+v\qquad\hbox{for some $v\in Y$}\cr \Rightarrow\quad&z=x+(y+v)\cr \Rightarrow\quad&z\in[x]\qquad\hbox{since $y+v\in Y$.}\cr}$$ Thus $[x+y]\subseteq[x]$. You can similarly show $[x]\subseteq[x+y]$ (try it), so $[x+y]=[x]$.
So in your problem, $$[x]=[\textstyle\sum a_i y_i + \sum b_k x_k]=[\sum b_k x_k]\ ,$$ and the rest is as in your post.