Consider a partition $a = x_0 < x_1 < \ldots < x_n = b$ and for a subinterval $I_j = [x_{j-1},x_j]$ define
$$D_{j}(g) = \sup_{x \in I_j}g(x) - \inf_{x \in I_j}g(x) = \sup_{x,y \in I_j}|g(x) - g(y)| \\ D_{j}(f \circ g ) = \sup_{x \in I_j}f(g(x)) - \inf_{x \in I_j}f(g(x)) = \sup_{x,y \in I_j}|f(g(x)) - f(g(y))|$$
Since $f$ is continuous it is uniformly continuous and bounded (by extension if necessary) on a closed interval $[c,d]$ such that $g([a,b]) \subset [c,d].$
Hence, $|f(x)| \leqslant M$ for $x \in [c,d]$ and for every $\epsilon >0$ there exists $\delta > 0$ such that if $|x_1 - x_2| < \delta$ then $|f(x_1) - f(x_2)| < \epsilon/(2(b-a))$ .
Since $g$ is integrable, if the partition norm $\|P\|$ is sufficiently small we have
$$U(P,g) - L(P,g) = \sum_{j=1}^n D_j(g) (x_j - x_{j-1}) < \frac{ \delta \epsilon}{4M}.$$
We can split the upper-lower sum difference $U(P,f \circ g) - L(P, f \circ g)$ into two sums as given by
$$\tag{1}U(P,f \circ g) - L(P, f \circ g) = \sum_{D_j(g) \geqslant \delta} D_j(f \circ g)(x_j - x_{j-1}) + \sum_{D_j(g) < \delta} D_j(f \circ g)(x_j - x_{j-1})$$
In the second sum on the RHS of (1) we have $D_j(f \circ g) < \epsilon/(2(b-a)$ since by uniform continuity $D_j(g) < \delta \implies |g(x) - g(y)| < \delta \implies |f(g(x)) - f(g(y))| < \epsilon/(2(b-a)$ for all $x,y \in I_j$.
Thus,
$$\tag{2}\sum_{D_j(g) < \delta} D_j(f \circ g)(x_j - x_{j-1}) < \frac{\epsilon}{2}$$
Considering the first sum on the RHS of (1), first note that
$$\sum_{D_j(g) \geqslant \delta} (x_j - x_{j-1}) \\ < \delta^{-1}\sum_{D_j(g) \geqslant \delta} D_j(g)(x_j - x_{j-1}) < \delta^{-1} [U(P,g) - L(P,g)] < \delta^{-1} \frac{\delta \epsilon}{4M} = \frac{\epsilon}{4M} .$$
Hence,
$$\tag{3}\sum_{D_j(g) \geqslant \delta} D_j(f \circ g)(x_j - x_{j-1}) < \sum_{D_j(g) \geqslant \delta} 2M(x_j - x_{j-1}) < \frac{\epsilon}{2}.$$
From (1), (2) and (3) we obtain
$$U(P,f \circ g) - L(P, f \circ g) < \epsilon,$$
and conclude that $f \circ g$ is integrable.
Best Answer
For the reverse implication we can construct a sequence of partitions $(P_n)_n$ and Riemann sums $(S(P_n,f))_n$ where both $\|P_n\| \to 0$ and the Riemann sums form a Cauchy sequence.
Hence, there exists $I \in \mathbb{R}$ and $N \in \mathbb{N}$ such that $S(P_n,f) \to I$ as $n \to \infty$, and for any $\epsilon > 0$ there exists $N \in \mathbb{N}$ such that for all $n \geqslant N$ we have
$$\|P_n\| < \delta, \quad |S(P_n,f) -I| < \frac{\epsilon}{2} $$
In the above, we could have taken $\delta$ such that $|S(P_1,f) - S(P_2,f)| < \frac{\epsilon}{2}$ when $\|P_1\|, \|P_2\| < \delta$. If $P$ is any partition where $\|P\| < \delta$, then since we already have $\|P_N\| < \delta$, it follows that
$$|S(P,f) - I| \leqslant |S(P,f) - S(P_N,f)|+ |S(P_N,f) - I| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon,$$
and $f$ must be Riemann integrable.
Construction of the sequences
There exists $\delta_1$ such that if $\|P\|, \|Q\| < \delta_1$, then $|S(P,f) - S(Q,f)| < 1$. We can assume $\delta_1 < 1$ and let $P_1$ be any partition with $\|P_1\| < \delta_1$ and $S(P_1,f)$ be any corresponding Riemann sum.
Further, there exists $\delta_2 < \min(\frac{1}{2},\delta_1)$ such that if $\|P\|, \|Q\| < \delta_2$, then $|S(P,f) - S(Q,f)| < \frac{1}{2}$. Let $P_2$ be any partition with $\|P_2\| < \min(\delta_2, \|P_1\|)$ and $S(P_2,f)$ be any corresponding Riemann sum.
Proceeding, for any $n \in \mathbb{N}$, there exists $\delta_n < \min(\frac{1}{n}, \delta_{n-1}) $ along with a partition $P_n$ and Riemann sum $S(P_n,f)$ such that if $\|P\|, \|Q\| < \delta_n$, then $|S(P,f) - S(Q,f)| < \frac{1}{n}$ and $\|P_n\| < \min( \delta_n , \|P_{n-1}\|)$.
Try to finish yourself by proving that $(S(P_n,f))$ is a Cauchy sequence.