Let: $v(s) = \exp\biggl({-}\int_a^s\beta(r)\,\mathrm{d}r\biggr)\int_a^s\beta(r)u(r)\,\mathrm{d}r,\qquad s\in I.$
Using the product rule, the chain rule, the derivative of the exponential function and the fundamental theorem of calculus, we obtain for the derivative
$[\exp(-\int_a^s\beta(r)\,dr)(\int_a^s\beta(r)u(r)dr)]'=$
applying $(uv)'=u'v+uv'$
$(e^{-\int_a^s\beta(r)\,dr})'(\int_a^s\beta(r)u(r)dr)+(e^{-\int_a^s\beta(r)\,dr})(\int_a^s\beta(r)u(r)dr)'=$
Chain rule on the first summand exp term, Newton Leibniz formula to the right term
$(e^{-\int_a^s\beta(r)\,dr})(-\int_a^s\beta(r)\,dr)'(\int_a^s\beta(r)u(r)dr)\quad+\quad(e^{-\int_a^s\beta(r)\,dr})(\beta(s)u(s)-\beta(a)u(a))=$
Newton Leibniz formula for the derivative of integral on the first summands factor
$(e^{-\int_a^s\beta(r)\,dr})(-\beta(s) + \beta(a))(\int_a^s\beta(r)u(r)dr)\quad+\quad(e^{-\int_a^s\beta(r)\,dr})(\beta(s)u(s)-\beta(a)u(a))=$
pulling out exp term
$(e^{-\int_a^s\beta(r)\,dr})[(-\beta(s) + \beta(a))(\int_a^s\beta(r)u(r)dr)\quad+\quad\beta(s)u(s)-\beta(a)u(a)]=$
$\beta(s),\beta(a)$ out
$(e^{-\int_a^s\beta(r)\,dr})[\beta(s)(u(s)-\int_a^s\beta(r)u(r)dr)+\beta(a)(-u(s)+\int_a^s\beta(r)u(r)dr)]$
However, wiki has no term with $\beta(a)(-u(s)+\int_a^s\beta(r)u(r)dr)$
$v'(s) = \biggl(\underbrace{u(s)-\int_a^s\beta(r)u(r)\,\mathrm{d}r}_{\le\,\alpha(s)}\biggr)\beta(s)\exp\biggl({-}\int_a^s\beta(r)\mathrm{d}r\biggr), \qquad s\in I,$
are there any assumptions or given conditions so that the second summand vanishes?
You applied the Leibniz integral rule incorrectly as mentioned in the comments. We are differentiating wrt to $s$, so $\frac{d}{ds}a=0$. Then we have
$$\left(e^{-\int_a^s\beta(r)\,dr}\right)'\left(\int_a^s\beta(r)u(r)dr\right)+\left(e^{-\int_a^s\beta(r)\,dr}\right)\left(\int_a^s\beta(r)u(r)dr\right)'$$ $$=e^{-\int_{a}^{s}\beta(r)dr}\left(\int_a^s\beta(r)u(r)dr\right)\frac{d}{ds}\left(-\int_{a}^{s}\beta(r)dr\right)+e^{-\int_{a}^{s}\beta(r)dr}\left(\frac{d}{ds}\int_{a}^{s}\beta(r)u(r)dr\right)$$ $$=e^{-\int_{a}^{s}\beta(r)dr}\left(\int_a^s\beta(r)u(r)dr\right)\left(-\beta(s)\right)+e^{-\int_{a}^{s}\beta(r)dr}\left(\beta(s)u(s)\right)$$ $$=\beta(s)e^{-\int_{a}^{s}\beta(r)dr}\left(u(s)-\int_{a}^{s}\beta(r)u(r)dr\right)$$