I would like to consider two PDE
$$\frac{\partial V}{\partial u}=\frac{\partial^{2} V}{\partial x^{2}}$$
such that
$$V(u,0)=\delta(u),V(0,x)=0,Domain=\{u>0,x>0\}$$
and
$$\frac{\partial V}{\partial u}=\frac{\partial^{2} V}{\partial x^{2}}$$
such that
$$V(0,x)=\delta(x),V(u,0)=0,Domain=\{u>0,x>0\}$$ equivalent? It seems that both means that $V(u,x)=0$ when $u>0,x>0$ and has the same physical meaning.
The two pde's are not equivalent. Physically speaking the first one $$\left\{\begin{matrix}\partial_t V = \partial_{xx}^2 V\\ V(0,t) = 0\\ V(x,0)=\delta(x)\end{matrix}\right.$$ means that you want to study some semi-infinite rod that has, as you said,an insulated end, to which you give at the start a unit impulse. The second one $$\left\{\begin{matrix}\partial_t V = \partial_{xx}^2 V\\ V(0,t) = \delta(t)\\ V(x,0)=0\end{matrix}\right.$$ means that you have the same semi-infinite rod to which, at any moment, you give a unit impulse at one and that at the start, all the rod is insulated but it is so just at the start along all the length of the rod!
The first problem seems more resonable to me and can be easily solved by taking the odd extension of the initial conditions and solve the heat equation for all $\mathbb{R}$. But from what I recall the odd extension of the Dirac delta is zero everywhere. You can see it even using the Green's function to evaluate the solution: the Green's function for this problem is $$G(x,x',t) = \frac{1}{\sqrt{4\pi t}}\left(e^{\frac{(x-x')^2}{4t}}-e^{\frac{(x+x')^2}{4t}}\right)$$ for which $$u(x,t) = \int_0^\infty G(x,x',t)\delta(x')dx' = 0 $$ Conceptually speaking, giving a unite impulse on an end which has to be zero all the time won't give anything to the rest of the rod. It's like having an immovable, indestructible wall and you hit it with a running car: the wall won't feel nothing at all. But, on the other hand, for your problem, you could give a unit impulse very close to $x=0$ using an initial condition such as $u(x,0)=\delta(x-\epsilon)$ for a small $\epsilon$