How should I think about potential energy surfaces for reactions of complex systems?
In order to understand what the PES really is, it is useful to see where this concept comes from.

A system of $M$ nuclei and $N$ electrons should be described by a multidimensional wavefunction $\Psi(\vec{r}_1,\dots,\vec{r}_N,\vec{R}_1,\dots,\vec{R}_M)$, where $\vec{r}$ denotes electronic coordinates and $\vec{R}$ denotes nuclear coordinates (we neglect spin for simplicity). Within the Born-Oppenheimer approximation, the total wavefunction is split in an electronic part $\phi$ and a nuclear part $\chi$:
$$
\Psi(\vec{r}_1,\dots,\vec{r}_N,\vec{R}_1,\dots,\vec{R}_M) = \phi(\vec{r}_1,\dots,\vec{r}_N;\vec{R}_1,\dots,\vec{R}_M)\chi(\vec{R}_1,\dots,\vec{R}_M)
$$
where the electronic wavefunction depends parametrically on nuclear positions (as it will be clear in a moment).

If you inject the Born-Oppenheimer ansatz in Schrödinger equation for a system of interacting nuclei and electrons and you integrate out electronic degrees of freedom you get an equation for the nuclear wavefunction only
$$
\left[ -\sum_I \frac{1}{2M_I} \nabla^2 + \sum_{I&lt;J}  \frac{Z_IZ_J}{|\vec{R}_I-\vec{R}_J|} + E(\vec{R}_1,\dots,\vec{R}_M) \right]\chi = E_\text{tot} \chi
$$
where $E(\vec{R}_1,\dots,\vec{R}_M)$ is the solution of the electronic problem
$$
\left[ -\sum_i \frac{1}{2}\nabla_{i}^2 +   \sum_{i&lt;j} \frac{1}{|\vec{r}_i-\vec{r}_j|} -\sum_{i,I} \frac{Z_I}{|\vec{r}_i-\vec{R}_I|} \right]\phi = E(\vec{R}_1,\dots,\vec{R}_M)\phi
$$
As you can easily see the solution of the electronic problem (both the energy and the wavefunctions) depend parametrically on nuclear coordinates. The potential energy surface is the potential energy appearing in the equation of the nuclear wavefunction, i.e.
$$
U(\vec{R}_1,\dots,\vec{R}_M)=E(\vec{R}_1,\dots,\vec{R}_M) + \sum_{I&lt;J}  \frac{Z_IZ_J}{|\vec{R}_I-\vec{R}_J|}.
$$
It is clear that finding the PES is a formidable task, since you have to solve the electronic problem for each possible set of nuclear coordinates. So, yes the PES is a multi-dimensional function of nuclear coordinates. If you have $M$ nuclei, your PES will be a function of at least $3M-6$ coordinates (once you remove translation and rotation of the whole system).

Since a multi-dimensional function is impossible to visualize, usually we study the PES by considering few interesting degrees of freedom (as the O-C distance and O-C-I angle in your example). Therefore, as you say, you can think as if you are exploring the PES locally and in a reduced space.



ADDITIONAL INFORMATIONS

Since the nuclear wavefunction is sharply picked, (heavy) nucley can usually be considered as classical particles. Thus instead of the nuclear quantum Hamiltonian
$$
H_n = -\sum_I \frac{1}{2M_I} \nabla^2 + \sum_{I&lt;J}  \frac{Z_IZ_J}{|\vec{R}_I-\vec{R}_J|} + E(\vec{R}_1,\dots,\vec{R}_M)
$$
you can consider its classical counterpart
$$
H_n^\text{cl} = \sum_I \frac{\vec{P}_I^2}{2M_I} + \sum_{I&lt;J}  \frac{Z_IZ_J}{|\vec{R}_I-\vec{R}_J|} + E(\vec{R}_1,\dots,\vec{R}_M) = \sum_I \frac{\vec{P}_I^2}{2M_I} + U(\vec{R}_1,\dots,\vec{R}_M)
$$
where $\vec{P}_I$ is the momentum of the nucleus $I$. From the laws of classical mechanics you can easily verify that the force acting on nucleus I is simply given by
$$
\vec{F}_I = -\nabla_{\vec{R}_I} U(\vec{R}_1,\dots,\vec{R}_M).
$$
So even if you consider the nuclei as classical particles, the PES has the same meaning, i.e. is the potential felt by the nuclei. Here you can easily apply a classical picture. A transition state is simply a state along your reaction coordinate where you have a local maxima of the PES for the reduced coordinates you are considering(as the C-O bond in you example).

I never looked to this into details but I think computational programs climb the PES uphill from the initial and final state of your system (i.e. they follows -$\vec{F}_I$) in order to find the TS.