Why is the surface of a torus flat? Why is the surface of a torus is said to be flat? If you consider the geometry of the torus, its surface has locally positive (spherical), negative (hyperbolic) and flat curvature.
 A: The metric on a circle $S^1$ is unique, it is always flat and hence its Riemannian curvature always vanishes.
However, there is a family of metrics on the torus $T^2 := S^1 \times S^1$. One of these is a flat metric. All these metrics are intrinsic
However, the flat metric in a sense has a claim to being an intrinsic metric and the usual metric in a sense is an extrinsic metric.
The torus can be defined as the product of two circles:

$T^2 := S^1 \times S^1$

Since the Riemann curvatures of both factors vanishes (since the curvature of any circle vanishes), the Riemann curvature of the torus vanishes too. And this gives the flat metric on the torus. Thus the flat metric turns up in a natural way without any embedding in any space being considered, either for the circular factors or for the torus itself
It is also possible to embed the smooth torus smoothly into 3-space and this latter space has a metric which we can induce onto the 3-torus. Thus this metric is naturally induced extrinsically, even though it is again an intrinsic metric.
In fact, the flat torus can be embedded into 4-space in such a way that its metric is induced from the metric of 4-space. This embedding is called the Clifford torus. Furthernore, it is the metric of the rectangular torus. This is the torus constructed from a rectangle by identifying opposite sides and we can see directly here the metric is flat because the rectangle is flat.
A: Not every torus is flat.
A 2-dimensional torus is any topological space which is homeomorphic (topologically equivalent) to a product of two circles.  However, knowing the topology is not enough to give you curvature information:  you need to specify a Riemannian metric on the torus if you want geometric data like this.  A metric is additional data on top of the topological structure.
The torus you are most familiar with can be obtained by rotating a circle around a line.  This torus (a "doughnut") inherits a Riemannian metric from the ambient space it is embedded in.  With this metric, this torus is not flat:  as you observe it has both regions of both positive and negative curvature.
Some tori are flat though!  The easiest example is the PacMan universe:  a square where if you exit through one side, you appear on the other side.  In other words, the left edge has been "identified" with the right edge, and the top edge has been "identified" with the bottom edge.  However, as far as PacMan is concerned his universe is not curved.  To be precise, when he parallel transports a vector around a tiny loop, he gets back exactly the same vector he started with.  This flat torus cannot be embedded in $\mathbb{R}^3$, but it can be embedded in $\mathbb{R}^4$ as Mozibur describes in their answer.
The Gaussian curvature is "intrinsic": it can be calculated just from the metric. More intuitively, an ant living on a donut embedded in 3D space could tell that his home was curved: they could verify that the angle sum theorem is not true to within a first order approximation, or that parallel transport of vectors changes the vectors. None of this depends on awareness of the embedding.  The "extrinsic" curvatures here are the "principle curvatures", which the ant cannot say anything about.
A: The usual 2-dimensional torus has positive Gaussian curvature on the outside, and negative on the inside. As simple as that.
https://en.wikipedia.org/wiki/Gaussian_curvature
