My line of thought was to try using Cauchy's formula to represent the values of the derivatives through the values of the functions themselves
That's the right idea.
I'm also a bit confused why the requirement is convergence only on compact subsets and not on the entire ball.
In general, the series/sequence of the derivatives will not converge uniformly on the entire disk, even when the series/sequence of functions does. Consider for a trivial example
$$\sum_{n=1}^\infty \frac{z^n}{n^2}.$$
The series converges uniformly on the closed unit disk, but the series of derivatives,
$$\sum_{n=1}^\infty \frac{z^{n-1}}{n}$$
does not, as the limit function is unbounded for $z\to 1$.
Cauchy's integral formula for the derivatives gives you an estimate
$$\begin{align}
\lvert f'(z) \rvert &= \left\lvert \frac{1}{2\pi i}\int_{\lvert z\rvert = \rho} \frac{f(\zeta)}{(\zeta-z)^2}\,d\zeta\right\rvert\\
&= \frac{1}{2\pi} \left\lvert \int_0^{2\pi} \frac{f(\rho e^{i\varphi})}{(\rho e^{i\varphi}-z)^2} \rho e^{i\varphi}\,d\varphi\right\rvert\\
&\leqslant \frac{\rho}{2\pi} \int_0^{2\pi} \frac{\lvert f(\rho e^{i\varphi})\rvert}{\lvert \rho e^{i\varphi}-z\rvert^2}\,d\varphi\\
&\leqslant \frac{\rho}{(\rho - \lvert z\rvert)^2}\cdot \max_{\lvert \zeta\rvert = \rho} \lvert f(\zeta)\rvert.
\end{align}$$
That estimate grows to infinity as $\lvert z\rvert \to \rho$, but it gives a finite uniform bound on every disk of smaller radius than $\rho$. Thus on every disk $D_{r_1}(0)$ with $r_1 < r$, for every $r_1 < \rho < r$, we have a uniform bound
$$\lvert f'(z)\rvert \leqslant \frac{\rho}{(\rho - r_1)^2} \cdot \sup_{\lvert \zeta\rvert \leqslant \rho} \lvert f(\zeta)\rvert$$
valid for all holomorphic functions on $D_r(0)$. For fixed $r_1$ and $\rho$, the factor with which the supremum of the moduli of $f$ is multiplied is constant, and thus uniform convergence of a series/sequence $f_n$ of functions holomorphic on $D_r(0)$ implies the uniform convergence of the series/sequence of the $f_n'$ on the smaller disk $D_{r_1}(0)$. Since every compact subset of $D_r(0)$ is contained in such a smaller disk, that is the locally uniform or compact convergence of the series/sequence of the derivatives.
Yes, we can be sure of that. Note that a compact subset $K$ of the disc is contained in $\{z: |z| \le r\}$ for some $r < 1$ (because the continuous real-valued function $|z|$ attains a maximum on a compact set). Take $r < s < 1$. Now $|a_n| s^n$ must be bounded in order for $\sum_n a_n s^n$ to converge, so there is $B$ such that $|a_n z^n| \le B (|z|/s)^n \le B (r/s)^n$ and the series converges uniformly on $K$.
Best Answer
What you did seems correct.
An alternative way is the following: let $M_R:=\sup_{|z|\leq R}|f(z)|$. For a fixed $R<1$, define $g(z):=\frac 1{1+M_R}f(Rz)$, from the open unit disk to itself. Then by Schwarz lemma, we have $|g(z)|\leq |z|$ for all $z\in D$, hence $|f(Rz)|\leq (1+M_R)|z|$ for all $z\in D$. We get that $|f(z)|\leq \frac{1+M_R}R|z|$ for all $z$ in the closed ball of center $0$ and radius $R$, which shows that the series $\sum_nf(z^n)$ is normally convergent on this set.
More simply, for a fixed $R\in (0,1)$, we have $$\sup_{|z|\leq R}|f(z^n)|\leq \sup_{|z|\leq R^n}|f(z)|.$$ Then we use continuity at $0$ of $f$.