Starting from $\vec{\nabla}\times\vec{E} = -\partial \vec{B}/\partial t$, take a surface integral of both sides to find
\begin{align} \iint_{S(t)} \vec{\nabla}\times\vec{E}\cdot d\vec{S} &= \iint_{S(t)} -\frac{\partial \vec{B}}{\partial t}\cdot d\vec{S}\\ \therefore\quad \oint_{\partial S(t)} \vec{E}\cdot d\vec{l} &= -\frac{d}{dt} \iint_{S(t)} \vec{B}\cdot \hat{S}\, dS - \oint_{\partial S(t)}\vec{v}\times\vec{B}\cdot d\vec{S} \end{align}
where the LHS of the second line makes use of Stokes' theorem, while the RHS makes use of the Leibniz integral rule.
$\vec{B}\cdot \hat{S}=-B$ (magnetic field points into the page, while take area vector to point out) and $\iint_S dS = A(t) = l x(t)$ is the area of the (rectangular) surface enclosed by the loop, where $x(t)$ is the length of the sides that is varying with time.
So:
\begin{equation} \mathcal{E} = \oint_{\partial S(t)} (\vec{E}+\vec{v}\times\vec{B})\cdot d\vec{l} = -\frac{d(-Blx(t))}{dt} = Bl\frac{dx(t)}{dt} = Blv \end{equation}