Why are we interested in this kind of second variation? Short version: why are we interested in two parameter-variations in 
contrast to one-parameter variations?
Long version:
In the literature on the calculus of variations, (specifically on harmonic mappings between Riemannian manifolds) one usually finds the following notion of second variation:
Let $M,N$ be Riemannian manifolds, $\phi:M \to N$ harmonic. Denote by $E$ the Dirichlet's energy functional, i.e $$ E(\phi)= \int_M |d\phi|^2\operatorname{dVol}_{M}$$
 We look at a smooth variation $\phi_{s,t}:M \to N$ with two parameters $s,t$, $\phi_{0,0}=\phi$, i.e a map $(-\epsilon,\epsilon) \times (-\epsilon,\epsilon) \times M \to N$.
Define $V=\left. \frac{d}{dt}\right|_{t=0} \phi_{0,t},W=\left. \frac{d}{ds}\right|_{s=0} \phi_{s,0}$, and
$$ H_{\phi}(V,W):=\left. \frac{\partial^2}{\partial s \partial t} \right|_{t=s=0} E(\phi_{s,t})$$ 
My question: What is the motivation for defining the variation in such a way? 
All the applications I have seen were concerned only with analyzing $ \left. \frac{\partial^2}{\partial t^2} \right|_{t=0} E(\psi_{t})$ where $\psi_t$ is a one-parameter variation of $\phi$, satisfying $V=\left. \frac{d}{dt}\right|_{t=0} \psi_t$. (This is reasonable since the main issue is to determine the nature of $\phi$ as a critical point of $E$, i.e is it a local maximum,minmum or a saddle point).
Thus, I see no apparent reason for introducing the two-parameters when we can use only one parameter. This seems like an unneseccary complication.
Indeed the main definitions (like weakly stable etc) concern only
 $H_{\phi}(V,V)$ which is another way of saying** $ \left. \frac{\partial^2}{\partial t^2} \right|_{t=0} E(\psi_{t})$. Why do we bother to pass through the more general $H_{\phi}(V,W)$? Are there any advantages to that approach?
__
**Indeed, given $V$ and $\psi_t$ such that $V=\left. \frac{d}{dt}\right|_{t=0} \psi_t$, define $\phi_{s,t}=\psi_{s+t}$. Then $V=\left. \frac{d}{dt}\right|_{t=0} \phi_{0,t}=\left. \frac{d}{ds}\right|_{s=0} \phi_{s,0}$, and $H_{\phi}(V,V)=\left. \frac{\partial^2}{\partial s \partial t} \right|_{t=s=0} E(\phi_{s,t})=\left. \frac{\partial^2}{\partial s \partial t} \right|_{t=s=0} E(\psi_{s+t})=\left. \frac{\partial^2}{\partial t^2} \right|_{t=0} E(\psi_{t})$
 A: One reason why this can be desirable is to check if a critical point of the functional is a minimum, a maximum, or just a flex point. This can be interesting for example if you are considering curves in a Riemannian manifold and your functional $E[\phi]$ is the length of the curve. Then minima correspond to geodesics.
A: After some thought, I came up witht the following:
$(1)$ The values of $H(V,V)$ for all $V$ uniquely determines $H(V,W)$ for all $V,W$, since $H$ is symmetric ($H(V,W)=H(W,V)$). Explicitly
$$ H(V,W)=\frac{1}{2}\big(H(V+W,V+W)-H(V,V)-H(W,W)\big), $$
so at least conceptually, it should not be harder to compute $H(V,W)$ than $H(V,V)$.
$(2)$ Point $(1)$ seems contradictory with what we know about analysis of critical points of functions $\mathbb{R}^n \to \mathbb{R}$. In order to determine the nature of a critical point of $f$, one has to know all the entries of the Hessian, not just the diagonal. So how knowing only the "diagonal" values $H(V,V)$ suffices in our case?
The reason is that here we know $H(V,V)$ for any V. In the analogous situation for functions $f:\mathbb{R}^n \to \mathbb{R}$, this corresponds to knowing $\operatorname{Hess} f(v,v)=v^T \cdot \operatorname{Hess} f \cdot v$ for every $v \in \mathbb{R}^n$, hence it enables us to determine whether or not the hessian is positive-definite for instance.
Indeed, knowing only the diagonal elements of the hessian does not suffice to determine the nature of critical points, since this  information is only about $\operatorname{Hess} f(e_i,e_i)$ (where $e_i$ are the standard basis vectors of $\mathbb{R}^n$).
Thus, there are really two different meanings here of what is the "diagonal"....
