How to prove that $\frac{d}{dt} \left( \int_{a}^{s(t)}f(x,t)dx +\int_{s(t)}^{b}f(x,t)dx \right) = \frac{ds}{dt}(f_{a} - f_{b}) $.... let $s(t) \in (a,b)$ and the function $f(x,t)$ is continuous except at $x=s(t)$, how to prove 
$$\frac{d}{dt} \left( \int_{a}^{s(t)}f(x,t)dx +\int_{s(t)}^{b}f(x,t)dx \right) = \frac{ds}{dt}(f_{a} - f_{b}) + \int_{a}^{s(t)}\frac{\partial f(x,t)}{\partial t}dx +\int_{s(t)}^{b}\frac{\partial f(x,t)}{\partial t}dx.$$
where $$f_{a} = lim_{x \rightarrow s(t)^{-}} f(x,t)$$
and
$$f_{b} = lim_{x \rightarrow s(t)^{+}} f(x,t).$$
My argument is that we must evaluate the integral of continuous part of $f(x,t)$, and about the neighborhoud of the discontinuity. 
What i have done (following @Stella advice):
$$ lim_{\triangle t \rightarrow 0} \frac{\int_{a}^{s(t+\triangle t)}f(x,t+\triangle t)dx + \int_{s(t+\triangle t)}^{b}f(x,t+\triangle t)dx - \int_{a}^{s(t)}f(x,t)dx - \int_{s(t)}^{b}f(x,t)dx}{\triangle t} = $$ $$ lim_{\triangle t \rightarrow 0} \frac{\int_{a}^{s(t+\triangle t)-s(t)+s(t)}f(x,t+\triangle t)dx - \int_{a}^{s(t)}f(x,t)dx}{\triangle t}  + lim_{\triangle t \rightarrow 0} \frac{\int_{s(t+\triangle t)-s(t)+s(t)}^{b}f(x,t+\triangle t)dx - \int_{s(t)}^{b}f(x,t)dx}{\triangle t} = $$
$$  lim_{\triangle t \rightarrow 0} \frac{\int_{a}^{s(t+\triangle t)-s(t)}f(x,t+\triangle t)dx}{\triangle t} + lim_{\triangle t \rightarrow 0} \frac{\int_{s(t+\triangle t)-s(t)}^{b}f(x,t+\triangle t)dx}{\triangle t} +  \int_{a}^{s(t)}\frac{\partial f(x,t)}{\partial t}dx +\int_{s(t)}^{b}\frac{\partial f(x,t)}{\partial t}$$
 A: In the long set of equations, the second equality is not justified. Also, inserting the $-s(t) + s(t)$ into the upper and lower limits of the two integrals after the first equality is valid, but looks like the wrong direction to go.
The theorem claimed is not true unless you also assume that the two partial derivatives used in the result actually exist; it's possible that you need to assume they exist and are continuous on the regions over which they're being integrated, but I haven't checked that yet. 
I suggest that you take the first integral, in the long sequence of equations, and rewrite its integrand as 
$$
f(x, t + \Delta t) = f(x, t+ \Delta t) - f(x, t) + f(x, t).
$$
Then split the integral  into two parts, the integrals of 
$$
f(x, t+ \Delta t) - f(x, t)
$$
and  of 
$$
f(x, t).
$$
The integral of the second part will combine nicely with the third integral in your first line, in the form
$$
\int_a^{s(t + \Delta t)} f(x, t) ~dx  - \int_a^{s(t)} f(x, t) ~dx,
$$
which will then simplify to
$$
\int_{s(t)}^{s(t + \Delta t)} f(x, t) ~dx,
$$
which is an integral of $f$ taken over a very short interval. The mean value theorem for integrals then tells you that this simplifies to the length of the interval times the value of the integrand at some point of the interval. That should get you on your way. (Especially when you divide the interval-length by the $\Delta t$ that you've got in the denominator ... that'll lead to an $s'(t)$ term in the limit.)
Most of the remaining work is probably algebra, the mean value theorem for integrals, and some careful checking about when you can take limits inside an integral or swap the order of limits.  
A: Assume that the functions
$$F_a(x,t):=\cases{f(x,t)\quad&$(a\leq x<s(t))$\cr f_a(t) &$\bigl(x=s(t)\bigr)$\cr}$$
and
$$F_b(x,t):=\cases{f(x,t)\quad&$(s(t)<x\leq b)$\cr f_b(t) &$\bigl(x=s(t)\bigr)$\cr}\qquad\ ,$$
where $f_a(t)$ and $f_b(t)$ are defined in the question, can be extended as $C^1$-functions to neighborhoods $U$ and $V$ as shown in the figure below. This means that there is a a "rupture" $t\mapsto x=s(t)$ between the regimes of $F_a(\cdot,\cdot)$ and $F_b(\cdot,\cdot)$.

Then we have
$$Q(t):=\int_a^{s(t)}f(x,t)\ dx+\int_{s(t)}^b f(x,t)\ dx=\int_a^{s(t)}F_a(x,t)\ dx+\int_{s(t)}^b F_b(x,t)\ dx\ ,$$
where we now can apply Leibniz' rule "with extras" to both integrals on the right hand side. We obtain
$$Q'(t)=\int_a^{s(t)}{\partial F_a(x,t)\over\partial t}\ dx+F_a\bigl(x(s(t),t\bigr)s'(t)+\int_{s(t)}^b {\partial F_b(x,t)\over\partial t}\ dx-F_b\bigl(x(s(t),t\bigr)s'(t)\ .$$
Plugging in the definitions of $F_a(x,t)$ and $F_b(x,t)$ this becomes
$$Q'(t)=\int_a^{s(t)}{\partial f(x,t)\over\partial t}\ dx+\int_{s(t)}^b {\partial f(x,t)\over\partial t}\ dx+\bigl(f_a(t)-f_b(t)\bigr)s'(t)\ ,$$
as claimed.
