Dual basis and nondegenerate symmetric bilinear form

Suppose that we are given a basis $\{e_1,\ldots,e_n\}$ and its dual $\{f_1,\ldots,f_n\}$ with respect to a nondegenerate symmetric bilinear form B. Now are given a different basis $\{z_1,\ldots, z_n\}$ and its dual $\{t_1, \ldots, t_n\}$. We can write $z_i=c_{ij}e_j$ for a matrix $C$, and we can similarly write the new dual basis as a matrix $D$ times the old dual basis. I think that $D$ is supposed to be the inverse transpose of $C$, but I don't seem to be able to reach this conclusion by calculation.

Your expression $z_i=c_{ij}e_j$ doesn't really makes sense; what you probably mean is that $C$ is the matrix whose columns hold the coordinates of the $z_j$ on the basis $(e_1,\ldots,e_n)$, so $z_j=\sum_{i=1}^nc_{i,j}e_i$. Now by definition of a dual basis the functions $f_i$ compute the coordinates of a vector on the basis $(e_1,\ldots,e_n)$, so that $v=\sum_{i=1}^nf_i(v)e_i$ for all $v\in V$, and in particular $c_{i,j}=f_i(z_j)$. The columns of the matrix $D$ should similarly express the coordinates of $t_j$ on the basis $(f_1,\ldots,f_n)$, so $t_j=\sum_{i=1}^nd_{i,j}f_i$ in $V^*$. Now the dual basis is a symmetric concept, so the coordinate of any $\alpha\in V^*$ is given by application to$~e_i$, so that $\alpha=\sum_{i=1}^n\alpha(e_i)f_i$. In particular one has $d_{i,j}=t_j(e_i)$.
To see the relation with the matrix $C$, one needs to consider the inverse $A=C^{-1}$ which gives the coordinates of the $e_j$ on the basis $(z_1,\ldots,z_n)$, namely $e_j=\sum_{i=1}^na_{i,j}z_i$. By the same argument as above, the entries of $A$ satisfy $a_{i,j}=t_i(e_j)$. Comparing we see that $a_{i,j}=d_{j,i}$: one has $D=(C^{-1})^\top$.