A more "real-variable" style condition is this. Let me write $f_n(\alpha)$ instead of $f(n,\alpha)$, and $f(\alpha) = \lim_{n \to \infty} f_n(\alpha)$. Suppose in some closed interval $I$, $f$ and all $f_n$ are $C^2$, $f_n \to f$ pointwise, and all f_n'' are uniformly bounded. Then f_n' \to f' uniformly on $I$.
It suffices to prove this in the case $f = 0$ (in general, take $g_n = f_n - f$ which satisfies similar conditions to $f_n$ but converges to $0$, so if g_n' \to 0 we have $f_n' \to f$).
Let $\alpha, \beta$ be distinct members of $I$, and let $B$ be a uniform bound for f_n'' on $I$. By Taylor's theorem, f_n(\alpha) - f_n(\beta) = f_n(\alpha) + f_n'(\alpha) (\beta - \alpha) + f_n''(\xi_n) (\beta - \alpha)^2/2 for some $\xi_n \in I$. Write this as f_n'(\alpha) = \frac{f_n(\beta) - f_n(\alpha)}{\beta - \alpha} - f_n''(\xi_n) \frac{\beta - \alpha}{2} Given $\epsilon > 0$ and $\alpha$, take $\beta$ so that $0 < |\beta - \alpha| < 2\epsilon/(3 B)$. Take $n$ large enough that $|f_n(\alpha)| < |\beta - \alpha| \epsilon/3$ and $|f_n(\beta)| < |\beta - \alpha| \epsilon/3$. Then we have |f_n'(\alpha)| < \epsilon/3 + \epsilon/3 + \epsilon/3 = \epsilon. Since this works for all $\epsilon$, conclude that f_n'(\alpha) \to 0 as $n \to \infty$.