Lets say we have the following:
$$ \sum_{k=0}^\infty z^k \sum_{j=0}^k \frac{1}{j!(k-j)!} B_{k-j}^f(x) \frac{d^{j}}{dx^{j}}[a_k(x)] $$
Would it be correct to say that:
$$ \sum_{k=0}^\infty z^k \sum_{j=0}^k \frac{1}{j!(k-j)!} B_{k-j}^f(x) \frac{d^{j}}{dx^{j}}[a_k(x)] = \left(\sum_{k=0}^\infty \frac{z^k}{k!} \frac{d^k}{dx^k}[a_k(x)]\right) \left(\sum_{k=0}^\infty \frac{z^k}{k!} B_k^f(x)\right) $$
No, this isn't correct. Not unless $a_k(x)$ is independent of the variable of summation, k.
$$\sum_{k=0}^\infty c_k = (\sum_{k=0}^\infty a_k)( \sum_{k=0}^\infty b_k)$$
Then
$$c_k = \sum^k_{j=0} a_{k-j}b_{j} $$
In your case...
$$\sum^k_{j=0} \frac{1}{j!(k-j!)}B^f_{k-j}(x)\frac{d^j}{dx^j}[a_k(x)]z^k = \sum^k_{j=0} (\frac{B^f_{k-j}(x)z^{k-j}}{(k-j!)})(\frac{z^j}{j!}\frac{d^j}{dx^j}[a_k(x)])$$
This implies that
$$\frac{B^f_{k-j}(x)z^{k-j}}{(k-j)!} = a_{k-j}$$ $$\frac{z^j}{j!}\frac{d^j}{dx^j}[a_k(x)] = b_j$$
But, as you can see in the $b_j$ expression, there is still $a_k(x)$, not $a_j(x)$