It is a mere application of Itô's lemma, but I must admit that the notation may turn out to be confusing because of two reasons. The first one is the presence of two different times, namely the time $t$ measuring the evolution of the stochastic process and the term $T$, which appear to coincide after some step, while the considered bivariate functions still treat them separately. The second reason is, I guess, the ambiguous notation $\int_0^t \ldots \mathrm{d}W_u$, which means $\int_{W_0}^{W_t} \ldots \mathrm{d}W_u$ actually.
Let's rewrite the quantity $S(t)$ by taking those ambiguities into account. Firstly, let's recall that $F(t,T)$ is a stochastic process, where $t$ is the real "passing" time and $T$ is just a parameter. Let's denote the dependence of processes on the time $t$ by a subscript, in the same manner as $W_t$, hence $F_t(T)$. In the same spirit, one has $S_t = F_t(t)$, where the second $t$ is still treated as a parameter, which appears to be equal to $t$ "by coincidence". Thus, we understand that $F_0(t)$ is only an initial condition with respect to the process, even if it is dependent on $t$ as a parameter; in consenquence, $F_0(t)$ will be later differentiated with respect to $t$ as a mere function and not as a stochastic process; in other words, we will get its differential with the help of a partial derivative instead of Itô's lemma, even it is a random variable.
Moreover, $S_t$ is also a function of a Brownian motion, so that $S_t = f(W_t,t)$, where
$$
f(x,t) = F_0(t)\exp\left(-\frac{1}{2}\int_0^t \sigma_u(t)^2 \mathrm{d}u + \int_0^x \sigma_u(t) \mathrm{d}W_u\right).
$$
Then, Itô's lemma is applied as usual, i.e. $\mathrm{d}S_t = \left(\partial_tf(W_t,t) + \partial_xf(W_t,t) + \frac{1}{2}\partial_x^2f(W_t,t)\right)\mathrm{d}t + \partial_xf(W_t,t)\mathrm{d}W_t$. Since the variables appear inside the integrals too, we must use Leibniz integral rule, hence :
$$
\begin{array}{rcl}
\partial_tf(W_t,t)
&=& \displaystyle
F_0(t)\exp\left(-\frac{1}{2}\int_0^t \sigma_u(t)^2 \mathrm{d}u + \int_0^{W_t} \sigma_u(t) \mathrm{d}W_u\right) \cdot \left(\frac{\partial_tF_0(t)}{F_0(t)} - \frac{1}{2}\sigma_t(t)^2 - \int_0^t \sigma_u(t)\partial_t\sigma_u(t) \mathrm{d}u + \int_0^{W_t} \partial_t\sigma_u(t) \mathrm{d}W_u\right) \\
&=& \displaystyle
S_t\left(\partial_t\ln F_0(t) - \frac{1}{2}\sigma_t(t)^2 - \int_0^t \sigma_u(t)\partial_t\sigma_u(t) \mathrm{d}u + \int_0^{W_t} \partial_t\sigma_u(t) \mathrm{d}W_u\right)
\end{array}
$$
where the parentheses on the right contains the derivative of the argument of the exponential, while
$$
\partial_xf(W_t,t) = F_0(t)\exp\left(-\frac{1}{2}\int_0^t \sigma_u(t)^2 \mathrm{d}u + \int_0^{W_t} \sigma_u(t) \mathrm{d}W_u\right) \cdot \sigma_t(t) = S_t\sigma_t(t),
$$
and
$$\partial_x^2f(W_t,t) = S_t\sigma_t(t)^2$$ in consequence. After collecting all the terms, we get :
$$
\frac{\mathrm{d}S_t}{S_t} = \left(\partial_t\ln F_0(t) - \int_0^t \sigma_u(t)\partial_t\sigma_u(t) \mathrm{d}u + \int_0^{W_t} \partial_t\sigma_u(t) \mathrm{d}W_u\right)\mathrm{d}t + \sigma_t(t)\mathrm{d}W_t
$$
And hence this is the full derivation of the stochastic differential equation.
