The answer to all of your italicized questions is "basically yes."
To understand why, we'll examine explicitly how the zero-point energy $E_0$ of the simplest one-dimensional lattice field theory depends on the physical spacing $d$ between configuration degrees of freedom on the lattice and the overall spatial size $L$ of the lattice.
Setting the stage -- coupled oscillators
Consider the theory of a finite number $N$ of coupled harmonic oscillators in one spatial dimension. To be more concrete, let's examine the "loaded string," the discrete (lattice) model for a string in which the string under tension $T$ is considered to be comprised of $N$ point masses $m$ separated by a distance $d$. Both the classical and quantum versions of this model can be solved exactly with normal modes.
In the classical case, the motion of the system is written in terms of normal modes of certain frequencies $\omega_n$ whose expressions can be written down in closed form1;
$$
\omega_n = 2\sqrt{\frac{T}{md}}\sin\left(\frac{n\pi}{2(N+1)}\right).
$$
In the quantum case, the Hamiltonian can be written as a sum over creation and annihilation operators corresponding to the normal modes:
$$
H = \sum_{n=1}^N(a^\dagger_n a_n + \tfrac{1}{2})\hbar\omega_n
$$
From this expression, we can recognize the zero-point energy, the energy of the vacuum state in which no mode is excited, as
$$
E_0 = \sum_{n=1}^N\tfrac{1}{2}\hbar\omega_n.
$$
The continuum limit
In both the classical case and quantum case, we can take a continuum limit so that the model approaches a field theory. By starting with the lattice model and taking the limit, the physical origins of divergences present in the field theories becomes clear. To properly take the continuum limit and understand its physical motivations, it helps to rewrite it in terms of a set of parameters $L,\lambda,v$ in place of $T, m, N$. We define the total length $L$, linear mass density $\lambda$, and "wave speed" $v$ of the string by
$$
L = (N+1)\,d, \qquad \lambda = \frac{(N+1)\,m}{L}, \qquad v = \sqrt{T/\lambda}.
$$
Writing the normal mode frequencies in terms of these parameters gives
$$
\omega_n = \frac{2v}{d}\sin\left(\frac{n\pi d}{2L}\right).
$$
The limit in which the loaded string becomes a continuous string governed by a field theory, either classical or quantum, is that in which the lattice spacing $d$ and mass $m$ per particle go to zero while the length $L$, mass density $\lambda$, and wave speed $v$ remain fixed. Let's examine the zero-point energy of the string in terms of our $L,\lambda,v$ parameterization to get a feel for the physical meaning of certain divergences in the continuum limit:
$$
E_0 = \sum_{n=1}^{L/d-1}\frac{\hbar v}{d}\sin\left(\frac{n\pi d}{2L}\right).
$$
In the continuum limit of an infinitely long string, we simultaneously take $d\to 0$ and $L\to\infty$ so that in particular, $d/L\to 0$. However, for getting a good physical understanding, we consider what would happen if you were to perform these limits separately. We examine two cases:
- The string has fixed length with lattice spacing tending to zero; $d\to 0$.
- The string has fixed lattice spacing $d$ but a length tending to infinity; $L\to\infty$.
For both, it helps to focus on two quantities: the density of the normal mode frequencies on the real line as measured by the spacing $\Delta\omega = \omega_{n+1} - \omega_n$ between successive mode frequencies, and the size of the largest mode frequency $\omega_\mathrm{max} = \omega_{L/d-1}$. Notice that in either case above, the ratio $d/L$ tends to zero, and this allows us to derive simple, approximate expressions for $\Delta\omega$ and $\omega_\mathrm{max}$ exhibiting how they scale with $d$ and $L$. For the frequency spacing we have
\begin{align}
\Delta\omega
&= \omega_{n+1} - \omega_n \\
&= \frac{\hbar v}{d}\left[\sin\left(\frac{(n+1)\pi d}{2L}\right) - \sin\left(\frac{n\pi d}{2L}\right)\right] \\
&= \frac{\hbar v}{d}\left[\frac{\pi}{2}\frac{d}{L} + O(\tfrac{d}{L})^2\right],
\end{align}
and for the maximum frequency, we have
\begin{align}
\omega_\mathrm{max}
&= \omega_{L/d - 1} \\
&= \frac{\hbar v}{d}\sin\left[\left(\frac{L}{d}-1\right)\frac{\pi d}{2L}\right] = \frac{\hbar v}{d}\sin\left[\frac{\pi}{2}\left(1-\frac{d}{L}\right)\right] \\
&= \frac{\hbar v}{d}\left[1 + O(\tfrac{d}{L})^2\right].
\end{align}
For $d/L$ small, we obtain the scalings
\begin{align}
\Delta\omega \sim \frac{1}{L}\, \qquad \omega_\mathrm{max} \sim \frac{1}{d}
\end{align}
Interpretation and intuitive understanding
From the results above, we observe that the system size $L$ controls the spacing (density) of the mode frequencies on the real line, and the lattice spacing controls how high the frequencies go. When the system size $L$ is large, the mode frequencies become dense on the real line. When the lattice size is small, we consider modes up to much higher frequencies in the energy sum. This is essentially just Fourier analysis. If you want to represent a function on a large domain as a sum over trigonometric functions, you need closely-packed frequencies, and this leads to the Fourier integral. If you want to represent the very small scale behavior of a function, you need to include higher frequencies in your Fourier sum, and this leads to the Fourier series.
Returning to the question of divergences, we can now see that in the QFT context, increasing the spatial size of the system without bound can cause a divergence by making the frequencies become dense -- this is called an "IR divergence." Decreasing the lattice separation causes a divergence by making the energy sum go to arbitrarily high frequencies -- this is called a "UV divergence." If we were to eliminate the UV divergence by cutting off the energy sum at a certain frequency, this would be equivalent in this simple system to only considering degrees of freedom above a certain minimum distance scale (lattice spacing).
- See e.g. Classical Dynamics of Particles and Systems, Thornton and Marion, 5th ed., pp. 498-503
