You are right, equation (11) is derived mechanically from (7) (by taking the derivative wrt to $T$ and then combining is with (7)), and somehow they think that (13) can be obtained from (11) without remembering (7). Maybe by smartly integrating (note for example that $B_tB_T - BB_{tT}$ is the numerator of the derivative wrt to $t$ of fraction $B/B_T$) and using the boundary condition (I couldn't figure it out).
Of course, what we can do is solve the first-order linear equation in $t$ (7)
$$ B_t = a(t)B-1. $$
With the usual primitive functions:
$$ \alpha'(t) = a(t), \; \; \beta'(t) = -{\rm e}^{-\alpha(t)}, $$
the general solution to equation (7) is
$$ B(t,T) = c(T){\rm e}^{\alpha(t)} + {\rm e}^{\alpha(t)}\beta(t), $$
with $c(T)$ arbitrary function of $T$.
As $B(T,T)=0$, we must have:
$$c (T)= -\beta(T).$$
So:
$$ B(t,T) = -{\rm e}^{\alpha(t)} \left(\beta(T) - \beta(t)\right).$$
We can then easily check that this solution respects (13):
$$ B(0,T) = -{\rm e}^{\alpha(0)} \left(\beta(T) - \beta(0)\right) $$
$$ B(0,t) = -{\rm e}^{\alpha(0)} \left(\beta(t) - \beta(0)\right) $$
$$\partial B(0,t)/\partial t = -{\rm e}^{\alpha(0)}\beta'(t) = {\rm e}^{\alpha(0)} {\rm e}^{-\alpha(t)}$$
Edit: Note that (11) can be written as:
$$ (B_T)_t =\frac{1-B_t}{B}B_T $$
which is equivalent to
$$ (\ln B_T)_t = \frac{1-B_t}{B}. $$
At this point we need to remember from (7) that the right hand side is a function of $t$ only, $a(t)$, otherwise it's getting cumbersome to progress from here. The solution is
$$ B_T = {\rm e}^{\alpha (t) + \gamma (T)} $$
for $ \gamma (T)$ an arbitrary function of $T$. Integrating wrt to $T$, we get:
$$ B(t,T) = {\rm e}^{\alpha (t)} (\Gamma (T) + \eta (t)) $$
for $ \eta (t)$ an arbitrary function of $t$ and $\Gamma^\prime = {\rm e}^{\gamma}$.
Boundary condition $B(T,T)=0$ then forces:
$$\Gamma(T) = -\eta(T). $$
So,
$$B(t,T) = -{\rm e}^{\alpha(t)} \left(\eta(T) - \eta(t)\right).$$
One more time, noting that
$$ B_t = -{\rm e}^{\alpha(t)}a(t)\eta(T) + {\rm e}^{\alpha(t)}a(t) \eta(t) + {\rm e}^{\alpha(t)} \eta^\prime (t),$$
(7) then implies:
$$\eta(t) = \beta(t). $$