The Hamiltonian by definition follows from the Lagrangian, which does presuppose a curve (that minimizes the action). Isn't the existence of a curve always assumed a priori with the Hamiltonian, in this way?

The Hamiltonian is not defined as the Legendre Transform of the Lagrangian. And the Lagrangian does not presuppose a curve, it also is a field, but it isn't a field on phase space is a field on a combination of position and velocity space. You can start with a $2n+1$ or a $2n$ dimensional manifold and have a symplectic form of the $2n$ part and get a Hamiltonian field without ever having a curve or a Lagrangian. And you can get a Lagrangian by starting with a product of a configuration space and its velocity space. And the velocity space is the space in which the true velocity could lie, so the combination of the configuration and the velocity space is basically a scalar field on the tangent manifold of the configuration manifold with maybe some time too.

Actions are functionals. They are like fields, but instead of taking a point in phase space and giving a number or taking a point on something like the tangent manifold of the configuration space and giving a number the action instead takes a curve and give a number. Sure, terminology is abused and you use the same word action for the number as the function.

Just like I tried to distinguish the Hamilton experienced, $\mathcal{H}(t),$ from the $2n+1$ variable field $$(q_i,p_i,t)\mapsto\mathcal{H}(q_i,p_i,t)$$ or the many instantaneous fields $$(q_i,p_i)\mapsto\mathcal{H}(q_i,p_i,t_1),$$

$$(q_i,p_i)\mapsto\mathcal{H}(q_i,p_i,t_2),$$

$$(q_i,p_i)\mapsto\mathcal{H}(q_i,p_i,t_3).$$

But keep in mind, that just becasue your first introduction to a Hamiltonian or a Hamiltonian field, is via Lagrangian, doesn't mean that is the definition.

And what's tricky is that someone might give you the Lagrangian of a particular system and then define the Hamiltonian for that same system in terms of that specific Lagrangian. But they are just trying to point towards a specific Hamiltonian.

Hamiltonian Mechanics is a very specific subject that stands on its own. And is strongly related to symplectic forms on even dimensional symplectic manifolds (phase space).

Lagrangian mechanics is a very specific subject that also stands on its own. And it is strongly related to tangent manifolds.

Action is a functional and is strongly related to path integrals.

And all the subjects have connections. But a Hamiltonian is a field and so is a Lagrangian. They are defined on different spaces.

And you can use them to do pure field theories which don't have curves.

$$\frac{d\mathcal{H}}{dt}= \dot{q_i}\frac{\partial\mathcal{H}}{\partial q_i}+\dot{p_i}\frac{\partial\mathcal{H}}{\partial p_i}+\frac{\partial\mathcal{H}}{\partial t}$$

And physically this is related to how the Hamiltonian experienced, $\mathcal{H}(t)=\mathcal{H}(q_i(t),p_i(t),t),$ changes as you move around in phase space, compared to how the Hamiltonian field $$(q_i(t),p_i(t))\mapsto \mathcal{H}(q_i(t),p_i(t),t)$$ changes in time.

And as you pointed out $\frac{d\mathcal{H}}{dt}= \frac{\partial\mathcal{H}}{\partial t}$ means $\dot{q_i}\frac{\partial\mathcal{H}}{\partial q_i}+\dot{p_i}\frac{\partial\mathcal{H}}{\partial p_i}=0.$

Which means the flow in phase space is orthogonal to the gradient of the Hamiltonian. That the Hamiltonian experienced isn't changed by the flow in phase space, but solely by the change of the Hamiltonian field in time.

This is not the same as Hamilton's equations. Because it does not, for instance, imply Hamilton's equations. It is a consequence of Hamilton's equations.

$$\frac{d\mathcal{H}}{dt}= \dot{q_i}\frac{\partial\mathcal{H}}{\partial q_i}+\dot{p_i}\frac{\partial\mathcal{H}}{\partial p_i}+\frac{\partial\mathcal{H}}{\partial t}$$

And physically this is related to how the Hamiltonian experienced, $\mathcal{H}(t)=\mathcal{H}(q_i(t),p_i(t),t),$ changes as you move around in phase space, compared to how the Hamiltonian field $$(q_i,p_i)\mapsto \mathcal{H}(q_i,p_i,t)$$ changes in time.

And as you pointed out $\frac{d\mathcal{H}}{dt}= \frac{\partial\mathcal{H}}{\partial t}$ means $\dot{q_i}\frac{\partial\mathcal{H}}{\partial q_i}+\dot{p_i}\frac{\partial\mathcal{H}}{\partial p_i}=0.$

Which means the flow in phase space is orthogonal to the gradient of the Hamiltonian. That the Hamiltonian experienced isn't changed by the flow in phase space, but solely by the change of the Hamiltonian field in time.

This is not the same as Hamilton's equations. Because it does not, for instance, imply Hamilton's equations. It is a consequence of Hamilton's equations.

I did not here say that this implies Hamilton's equations and did say that it is a consequence of those equations. What I did ask here is whether there is a connection between the principles that are used to derive Hamilton's equations and the equality of the total and partial derivatives of the Hamiltonian.

Let's step way way back. You can have fields like $$(q_i,p_i)\mapsto \mathcal{H}_1(q_i,p_i)$$ and $$(q_i,p_i)\mapsto \mathcal{H}_2(q_i,p_i)$$ and $$(q_i,p_i)\mapsto \mathcal{H}_3(q_i,p_i).$$

Or even continuum many such fields parameterized by time. In other words for each time, $t,$ you get a field $$(q_i,p_i)\mapsto \mathcal{H}(q_i,p_i,t).$$

So a single function of $2n+1$ variables can be a time varying scalar field. Great, fields have partials. There are partials for each of the $2n+1$ variables:

$$\frac{\partial\mathcal{H}}{\partial q_1} ,\dots,\frac{\partial\mathcal{H}}{\partial q_n}, \frac{\partial\mathcal{H}}{\partial p_i},\dots, \frac{\partial\mathcal{H}}{\partial p_n},\frac{\partial\mathcal{H}}{\partial t}.$$

And fields only have partials, not total derivatives. How do you get a total derivative? You need a function of a single variable. You don't have one. You need one to be able to take a total derivative. If, in addition to this time changing field $(q_i,p_i,t)\mapsto\mathcal{H}(q_i,p_i,t)$ you had a curve $$t\mapsto (q_i(t),p_i(t))$$ then you can make a new function of a single variable $$\mathcal{H}(t)=\mathcal{H}(q_i(t),p_i(t),t).$$

This function could have a derivative (and it does if the field is smooth and the curve is differentiable). And since the field also has a partial derivative in time you can ask whether the total derivative (of the Hamiltonian experienced, which depends on the Hamiltonian and the curve) equals the partial time derivative when the latter (which is a field) is evaluated at the corresponding point on the curve.

So the property of the total derivative equaling the partial derivative is not a property of the field. And the field is the Hamiltonian, so it is not a property of the Hamiltonian.

Hamiltonian's equations give you some curves. And the curves they give you can be used to make single variable functions. Those single variable functions can satisfy the equation you provided if you evaluate the field on the right hand side at the appropriate place.

But the equation you started with began by assuming a curve of some kind. And some of the curves that satisfy the equation you started with do not satisfy Hamiltonian's equations.

There is no more connection than that. Physically your equation has the curve be instantaneously orthogonal to the instantaneous gradient of the instantaneous Hamiltonian (the $2n$ dimensional gradient of the field at that instant). Some curves do that without satisfying Hamilton's equations. The curves that satisfy Hamilton's equations do satisfy that.

But the equation presupposed a curve, and it is not a property of the Hamiltonian. And the definition of the Hamiltonian is a field of $2n+1$ variables. And Hamilton's equations relates curves to those fields in a particular way.

Finally, you did ask us to verify that

Hamilton's equations follow from the definition of the Hamiltonian

And they don't, the Hamiltonian is just a field, a function from $2n+1$ variables and doesn't even require any curves at all. So Hamilton's equations do not follow from the definition of a Hamiltonian.