Yes, a "flat" universe is infinitely large, and hence "non-circular". I think perhaps you are confusing a flat and "closed" geometry.
Flat universes
A flat universe is a universe where your good old high-school math works as expected: the angles of a triangle sum up to $\pi$ radians (180 degrees), the circumference of a circle is $\pi$ times its diameter, parallel lines never meet, etc. In a 2D analogy, this corresponds to the geometry of a table; hence the name "flat".
Closed and open universes
But "flat" isn't the only possible geometry of space. You may, in principle, imagine any crazy geometry you like. However, we like to believe that the Universe is homogeneous and isotropic, i.e. that it looks the same everywhere and in all directions. Indeed, it seems to be on sufficiently large scales. If this is true, then there are only two other possibilities, namely "closed" (or "spherical") and "open" (or "hyperbolic").
Non-flat geometries are difficult to visualize in 3D, but the 2D analogy of a closed universe is that of a sphere, like the surface of a ball. On the 2D spherical surface, the angles of a triangle sum up to more than 180º, the circumference of a circle is less than $\pi$ times its diameter, and lines that start out parallel eventually meet. Moreover, if you travel far enough, you will get back to where you started.
The same is true in 3D spherical space. You may even, in principle, see your own back, although that would require either 1) that you start looking exactly at the Big Bang, in which case the light from your back will reach exactly when the universe collapses in a Big Crunch, or 2) that you have a somewhat fine-tuned amount of dark energy to slow down the collapse, without making it expand too fast so the photons never reach you.
Measuring the curvature in theory
Visualizing the 2D sphere is easy when it's embedded in a 3D space. But to describe how large its curvature is, you actually don't need the embedding 3D. You can just do a geometric measurement and compare your result to what you'd expect in flat space.
The quantity usually used to describe the magnitude of the curvature is $K$, which for a 2D sphere equals $1/R^2$ where $R$ is the radius of the sphere. If you draw a triangle of area $A$, and measure its three angles $\theta_{1,2,3}$, then the curvature is given by
$$
K = \frac{\sum_i \! \theta_i - \pi}{A}.
$$
This formula is valid for spaces of any numbers of dimensions (higher than 1D).
For instance, consider a triangle going from Ecuador, to Gabon, to the North Pole, and back to Ecuador. In this case, the sum of the angles is 270º, or $3\pi/2$, and the area of the triangle is $1/8$ of Earth's surface area, i.e. $4\pi R_\mathrm{Earth}^2/8$. Hence, the curvature is
$$
K = \frac{3\pi/2 - \pi}{4\pi R_\mathrm{Earth}^2 / 8} = \frac{1}{R_\mathrm{Earth}^2}.
$$
There are several other mathematical ways to measure the curvature. For instance you can parallel transport a vector along two difference routes; only in a flat space will they end up pointing in the same direction. In the example above, transporting a vector pointing from Ecuador to Gabon along the shortest path (red path in the figure below), and transporting it via the North Pole (green path), gives a difference of 90º.
So, to see the curvature of a 2D space in a higher-dimensional, flat space requires that theis space be 3D. But as described above, seeing the curvature is not required in order to measure it.
Furthermore, to see the curvature of a 3D space in a higher-dimensional, flat space, actually 4D wouldn't be enough — you would need 6 dimensions. And to see a 4D space (e.g. spacetime), you'd need 10 dimensions. But just as in the 2D case, there's no need for it, and to my knowledge (but please correct me if I'm wrong) no-one claims the existence of such higher dimensions.
Measuring the curvature in practice
In practice, this it not how we measure the curvature of our Universe. Rather, we use the Friedmann equation which gives the relation between the expansion rate $H$ of the Universe, the density parameters $\Omega_i$ of its various components (dark matter, normal matter, dark energy, and radiation), and its curvature.
The density parameters are defined as the densities $\rho_i$, divided by the critical density $\rho_c$, which is the density needed for a flat universe. If the total density $\rho_\mathrm{tot}$ is greater than $\rho_c$ — i.e. if $\Omega_\mathrm{tot} \equiv \sum_i\Omega_i > 1$ — the universe is closed.
Although "curvature" is not really a component of the Universe (such as e.g. matter), it may be parameterized as a corresponding density, and thus, in the Friedmann equation, appear as a term $\Omega_K \equiv 1 - \Omega_\mathrm{tot}$:
$$
\frac{H^2(a)}{H_0^2} =
\frac{\Omega_\mathrm{r}}{a^4} +
\frac{\Omega_\mathrm{M}}{a^3} +
\frac{\Omega_K}{a^2} +
\Omega_\Lambda.
$$
Here, subscripts $0$ refer to the values measured today, whereas in general the values are measured at a time when the scale factor — the "size" of the Universe — is $a$. Subscripts $r$, $M$, and $\Lambda$ denote radiation, matter (dark+normal), and dark energy.
Hence, measuring the densities and expansion rates (preferably at various epochs given by $a$) will tell you $K$. This can be done in several different ways, e.g. by observing supernova brightnesses, the cosmic microwave background, the large-scale structure of the Universe, gravitationally lensed galaxy clusters, etc.
Luckily, the various more or less independent methods return roughy the same answer. And this answer seems to imply that our Universe is very close to flat; within the margins of the errors, it is consistent with being exactly flat (Planck Collaboration et al. 2018). Hence, for most practical purposes we may regard the Universe as flat. There was, however, recently a paper by di Valentino et al. (2019) claiming evidence for a positively curved (i.e. closed) Universe. But they considered only one out of Planck's many data sets, so it seems to me a bit like cherry-picking.
