I know little about how the path integral in quantum field theory works
This is not good news, but I will try give an answer anyhow. Let us consider gauge theory, with gauge group $G$, in four dimensions and in Euclidean time. Here the gauge field is a $G$-connection one-form $A$, which is a fancy way to say the following
- $A_\mu$ transforms as a covector under spacetime coordinate transformations (in particular Lorentz transformations). $A_\mu$ are the components of the one-form $A = A_\mu \mathrm{d}x^\mu$. This is the meaning of the one-form part of the italic text above
- $A_\mu$ has values in the Lie algebra of $G$ (i.e. $A_\mu = A_\mu^a t_a$ for $t_a$ a basis of the Lie algebra) and under a gauge transformation $g(x)$ transforms as $A^\prime_\mu = gA_\mu g^{-1} - g^{-1}\partial_\mu g$. This is the meaning of $G$-connection
Then you pick an action $S[A]$, which is some functional of $A$ which is gauge invariant and you construct the partition function for the quantum theory
\begin{equation}
Z = \int \mathcal{D}A \exp(-S[A])
\end{equation}
This is not really the correct expression for the partition function as here we are integrating over values of $A$ which are gauge equivalent and some form of gauge fixing is required via, for example, BRST.
However what really matters here is what the integration really is. The path integral measure can be tought informally as
\begin{equation}
\mathcal{D}A = \prod_{x\in \mathbb{R^4}} \prod_{a=1}^{\dim \mathfrak{g}}\mathrm{d}A_1^a(x)\mathrm{d}A_2^a(x)\mathrm{d}A_3^a(x)\mathrm{d}A_4^a(x)
\end{equation}
You see it is a product over a continuous index, the position $x$ and a product over the Lie algebra index. It is also easy to see how the integration measure changes with the dimension of space, we could write
\begin{equation}
\mathcal{D}A = \prod_{x\in \mathbb{R^d}} \prod_{a=1}^{\dim \mathfrak{g}}\prod_{\mu=1}^d\mathrm{d}A_\mu^a(x)
\end{equation}
Now what happens when we go to dimension $d = 0$. Well $\mathbb{R}^0$ is just a single point i.e. $\mathbb{R}^0 = \{0\}$, thus product over the continuos index disappears, also $A$ does not depend on the space coordinates $x$ anymore since space is just one point and it is just an element of the Lie algebra $\mathfrak{g}$ (a "matrix") and the measure is
\begin{equation}
\mathcal{D}A = \prod_{a=1}^{\dim \mathfrak{g}}\mathrm{d}A^a
\end{equation}
which is just the measure of a $(\dim \mathfrak{g})$-dimensional integral. I guess you confusion may come from the fact that when $G = U(N)$ we have $\dim \mathfrak{u}(N) = N^2$, then in when $N \to \infty$ we get another infinite dimensional integral. However I think the catch here is that first for the large $N$ limit you consider a $U(N)$ theory with arbitrary $N$, do your computations and then look at what happens when $N$ is large. The path integral is an $N^2$-dimensional integral with finite $N$ when you compute it. Second the matrix theory path integral does not have the "product of differential over a continuous index", which does not make sense and needs to be made sense of somehow (like via a discretization of space). For matrix models there are no functional integrals anymore and the path integral is just a regular multidimensional integral
