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There are multiple issues with how you compute the exponent of $r$.

The first issue is you compute $r^{j-1}r^k = r^{j+k}$ rather than $r^{j+k-1}$.

You compound this error when you substitute $j = k+1$ into $r^{j+k}$ (which should be $r^{j+k-1}$) in the formula for $\left\langle\frac{\partial f}{\partial z},g\right\rangle$. You end up with $r^{2k+3}$ rather than $r^{2k+1}$ (which should actually be $r^{2k}$).

You make the same mistake in the formula for $\langle f,zg\rangle$ where $r^{j+k+1}$ should become $r^{2k+2}$ but you have $r^{2k+3}$. That said, in this case you make another mistake which cancels this out by "multiplying" $r^{2k+3}r$ to just $r^{2k+3}$ instead of $r^{2k+4}$ when you pull the sum out of the integral.

Without these mistakes you would end up with $$\begin{align}
\left\langle \frac{\partial f}{\partial z}, g\right\rangle &= \sum\_{k=0}\^\infty a\_{k+1}\overline{b\_k} (k+1)\int_0^\infty e^{-r\^2}r\^{2k+1} dr \\\\
\langle f, zg \rangle &= \sum_{k=0}^\infty a\_{k+1}\overline{b\_k} \int_0^\infty e^{-r^2}r^{2k+3} dr 
\end{align}$$

Applying integration by parts to the integral in the second equation will reduce it to the first, i.e. a $k+1$ factor will pop out and the power of the $r^{2k+3}$ will be reduced by $2$. (This is easiest to see, for me at least, by first doing a $u$-substitution with $u=-r^2$.)