Treating the Schrödinger equation as an ordinary differential equation How exactly can I interpret the equation in this form?

In order to develop a theory of solutions also in the general case, one interprets the time-dependent Schrödinger equation
$$
\mathrm{i}\frac{\mathrm d}{\mathrm dt}\psi(t)=H\psi(t)
$$
as an ordinary differential equation for the vector-valued function $\psi:\mathbb R_t\to\mathcal H := L^2(\mathbb R^{Nd})$. However, since $H$ is not a bounded linear operator on $\mathcal H$, this theory turns out to be much more subtle than the theory of linear ODEs on finite dimensional spaces. These considerations will lead us to the spectral theorem for unbounded self-adjoint operators on Hilbert spaces, one of the central mathematical results discussed in this course.

So we treat $\psi$ as a curve in $Nd$-dimensional space, thus obtaining a solution for $\psi(t)$, but how could we ignore all the position dependence?
 A: $\psi$ is a curve through $\mathcal H:=L^2(\mathbb R^{Nd})$, not through $\mathbb R^{Nd}$ itself.  That is, for each $t\in \mathbb R$ we have that $\psi(t)\in L^2(\mathbb R^{Nd})$ is loosely$^\ddagger$ a square-integrable function.
You must be careful to distinguish between $\psi$, which is a curve through $\mathcal H$, and $\psi(t)$, which is the element of $\mathcal H$ the curve passes through at time $t$.  $\psi$ is a function of one variable which eats a time $t$ and spits out the vector $\psi(t)$, which is itself a square-integrable function.  It wouldn't make any sense for $\psi$ itself to accept a position as an input - what would that even mean?  Instead, it is $\psi(t)$ - which is often interpreted as a function of position (i.e. the position-space wavefunction) - which is able to take position as an input variable.
To make this explicit, we might use notation like $\big[\psi(t)\big](\mathbf x)$, which makes it clear that $t\in \mathbb R$ is a number which we plug into $\psi$ to get a function $\psi(t)\in L^2(\mathbb R^{Nd})$, and $\mathbf x\in \mathbb R^{Nd}$ is a vector we plug into $\psi(t)$ to get a number $\big[\psi(t)\big](\mathbf x)\in \mathbb C$.  This notation is the stuff of nightmares, so I personally prefer the symbol $\psi_t(\mathbf x)$ instead.
In most of the pedagogical literature, authors tend to sweep this discussion under the rug and simply write $\psi(t,\mathbf x)$; however, my view is that this obscures the distinction between space and time which occurs in non-relativistic quantum mechanics, and leads to deep misunderstandings.
When we write down the Schrodinger equation, we need to interpret the terms carefully. In my preferred notation:

*

*$\psi$ is a vector-valued function of one variable, and $\psi'$ is its vector-valued derivative

*$\psi_t$ and $\psi'_t$ are the vectors in $\mathcal H$ which result from evaluating $\psi$ and $\psi'$ at time $t$

*$H$ is a linear operator which eats a vector and spits out another vector

$$\longrightarrow  i \hbar\psi'_t = H\big(\psi_t\big)$$

$^\ddagger$Really an equivalence class of functions, where we identify two functions $f$ and $g$ as the same element of $L^2(\mathbb R^{Nd})$ if $\int \mathrm d^{Nd}x |f(x)-g(x)|^2 = 0$.
