Dimension of a quotient space

Setup. Let $(e_1,\ldots,e_k)$ be a basis of $F$ and extend to a basis $(e_1,\ldots,e_k,e_{k+1},\ldots,e_n)$ of $E$.

Span of the quotient. Every vector $v\in E$ decomposes as $v=\sum_{i=1}^k a_i e_i + \sum_{j=k+1}^n b_j e_j$. Modulo $F$, the first sum vanishes, so $[v]=\sum_{j=k+1}^n b_j [e_j]$. Thus the classes $[e_{k+1}],\ldots,[e_n]$ span $E/F$.

Independence. Suppose $\sum_{j=k+1}^n b_j [e_j]=0$. Then $\sum b_j e_j\in F$, so it is a combination of the $e_1,\ldots,e_k$. By independence of the full basis, all $b_j=0$.

Conclusion. $([e_{k+1}],\ldots,[e_n])$ is a basis of $E/F$, so $\dim(E/F)=n-k=\dim E-\dim F$.