The functions $f_n$ being continuous, the fundamental theorem of calculus tells us that
$$
f_n(x) = f_n(x_0) + \int_{x_0}^xf_n'(t)\,dt
$$
for $x \in [a,b]$ (this is theorem 6.21 in Rudin). Suppose that the functions $\{f_n'\}$ converge uniformly on $[a,b]$ to a (continuous) function $h$. Then define
$$
f(x) := f(x_0) + \int_{x_0}^x h(t)\,dt.
$$
By theorem 7.16 in Rudin, we know that
$$
\int_{x_0}^x h(t)\,dt = \lim_{n \to \infty} \int_{x_0}^x f_n'(t)\,dt = \lim_{n \to \infty} (f_n(x) - f_n(x_0))
$$
where the last equality is again the fundamental theorem of calculus. This shows that indeed
$$
f(x) = \lim_{n \to \infty} f_n(x).
$$
But since $h$ is continuous (this is because we supposed the $f_n$'s to be of class $C^1$), we can again invoke the fundamental theorem of calculus to say that
$$
f(x_0) + \int_{x_0}^x h(t)\,dt
$$
is a differentiable function on $[a,b]$ with derivative h(x) (this is theorem 6.20 in Rudin). In other words, $f(x)$ is differentiable with $f'(x) = h(x) = \lim_{n \to \infty} f_n'(x)$, as we wanted.
I can conclude a subsequence of $f_n$ converges uniformly. But the existence of a uniformly convergent subsequence does not guarantee the original sequence converges uniformly.
Generally, that indeed does not imply the uniform convergence of the full sequence.
However, here we are in a special situation, since we know that the full sequence converges pointwise, and an equicontinuous sequence that converges pointwise converges uniformly on compact sets.
Since the sequence in question is not only equicontinuous but equilipschitz, the proof is easier:
Let $\varepsilon > 0$ be given. Since $\lvert f_n'(x)\rvert \leqslant M$ for all $n$ and $x\in [a,b]$, we have
$$\lvert f_n(x) - f_n(y)\rvert \leqslant \frac{\varepsilon}{4}$$
for all $x,y\in [a,b]$ with $\lvert x-y\rvert \leqslant \frac{\varepsilon}{4M}$.
Choose $N$ large enough that $\frac{b-a}{N} < \frac{\varepsilon}{4M}$, and let $x_k = a + k\frac{b-a}{N}$ for $0 \leqslant k \leqslant N$. For each $k$, there is an $n_k \in \mathbb{N}$ such that $\lvert f_n(x_k) - f(x_k)\rvert < \frac{\varepsilon}{4}$ for all $n \geqslant n_k$. Let $n(\varepsilon) = \max \{ n_k : 0 \leqslant k \leqslant N\}$.
Then, for every $x \in [a,b]$ and $n, m \geqslant n(\varepsilon)$ we have
$$\begin{align}
\lvert f_n(x) - f_m(x)\rvert &\leqslant \lvert f_n(x) - f_n(x_k)\rvert + \lvert f_n(x_k) - f(x_k)\rvert + \lvert f(x_k) - f_m(x_k)\rvert + \lvert f_m(x_k) - f_m(x)\rvert\\
&\leqslant \frac{\varepsilon}{4} + \frac{\varepsilon}{4} + \frac{\varepsilon}{4} + \frac{\varepsilon}{4} = \varepsilon,
\end{align}$$
for $k = \left\lfloor N\frac{x-a}{b-a}\right\rfloor$, so $\lvert x-x_k\rvert \leqslant \frac{\varepsilon}{4M}$.
Thus, for every $\varepsilon > 0$, we found an $n(\varepsilon)\in\mathbb{N}$ with
$$\sup \{ \lvert f_n(x) - f_m(x)\rvert : x \in [a,b], \, n,m \geqslant n(\varepsilon)\} \leqslant \varepsilon,$$
i.e. the sequence converges uniformly.
Best Answer
Uniform convergence of $\sum f_n'$ is much stronger than what is needed to guarantee uniform convergence of $\sum f_n$. Uniform boundedness of the partial sums is sufficient.
Suppose $\sum_{n=0}^ \infty f_n(x)$ converges pointwise and
$$\tag{*} \left|\sum_{n=0}^m f'_n(x)\right| \leqslant B$$
for all $x \in (a,b)$ and all $m \in \mathbb{N}.$
Take a partition $a = x_0 < x_1 < \ldots x_r = b$ where $x_k - x_{k-1} < \epsilon/(2B)$. We have convergence of $\sum f_n(x)$ at the $r+1$ endpoints, and, hence, there exists a positive integer $N$ such that for all $p > m > N$ we have $$|g(x_k)| := \left|\sum_{n = m+1}^{p} f_n(x_k) \right| < \frac{\epsilon}{2} \quad (k = 0,1,\ldots,r)$$
For any $x \in (a,b)$ there is a closest point $x_k$ such that $|x-x_k| < \epsilon/(4B)$. By the reverse triangle inequality and MVT, we have for some $\xi$ between $x$ and $x_k$,
$$|g(x)| \leqslant |g(x_k)| + |g(x) - g(x_k)| = |g(x_k)| + |g'(\xi)||x- x_k| \leqslant \frac{\epsilon}{2} + \frac{\epsilon}{4b} |g'(\xi)|$$
The inequality (*) leads to $|g'(\xi)| = \left|\sum_{n = m+1}^{p} f'_n(\xi) \right| \leqslant 2B$ and it follows that for all $x \in (a,b)$,
$$|g(x)| = \left|\sum_{n=m+1}^{p}f_n(x)\right| < \epsilon$$