$(M^2,g)$ is Riemannian surface. $\lambda(t):[0,L]\rightarrow M^2$ is a simple closed curve. The variation of $\lambda (t)$ is $$ f(s,t):(-\epsilon,\epsilon)\times[0,L] \rightarrow M^2 $$ such that $f(0,t)=\lambda(t), f(s,0)=f(s,L)$. For fix $s\in (-\epsilon,\epsilon)$, $N(s,t)$ is the normal vector of $f(s,t)$. Then, I guess that the directed area closed by $f(0,t)$ and f(s,t) is $$ A(s)=\int_0^L\int_0^s N(\tau,t)\cdot f_s(\tau,t) d\tau dt \tag{1} $$ where $f_s(\tau,t)=\partial_s f(\tau,t)$.
Of course, the above statement is not rigorous, in order to guarantee the existence of normal vectors we need $f(s,t)$ to be regular enough. Besides, $\epsilon$ may needs to be sufficiently small. But I don't think that's the core. So I didn't really think about the rigor.
I guess the (1) in another problem. Seemly, I'm not the only one who think it's right. But now, when I try to show it is right, I don't know how to start...
I've drawn a picture of the problem, just to make it a little bit more intuitive
