Let's say I want to compute the distance between two far points on Earth, say Toronto and Brazil. I can do this by getting in my car, setting my odometer to zero and then driving to Brazil. For me, the distance between these two points is the number my odometer spits out.
An observer outside Earth would be using the metric $dx^2+dy^2+dz^2$ and would calculate a different distance between these two points. From what I know, the induced metric on the sphere is the round metric which is $r^2(d\theta^2+\sin\theta^2 d\phi^2)$. I'm not sure what this is in terms of $x,y,z$.
If the person outside the Earth were to put Toronto and Brazil into the round metric, would they get the same answer the odometer gave? Is the round metric suppose to be the metric a person on the sphere would use?
Suppose this person in Toronto takes herself as the origin of the $xy$-plane she lives on. Her metric would be $dx^2+dy^2$. If the person observing the Earth also put their origin at Toronto, would the round metric coming from $dx^2+dy^2+dz^2$ coincide with this i.e. $g_{\text{round}}=dx^2+dy^2$? The person on the Earth is calculating the length of this part of a great circle using the metric $dx^2+dy^2$ not the round metric? Are the round metric and the surface's metric the same objects; is there a difference between spitting out the same lengths and being the same thing?
The round metric on the sphere is the metric induced from the Euclidean metric $dx^2 + dy^2 + dz^2$ on $\mathbb{R}^3$ (in fancy language, the round metric is the pullback of the Euclidean metric under the inclusion of the sphere into $\mathbb{R}^3$). This means in particular that if the person "outside the earth" decided to measure the length of a path on the surface of the earth (using the Euclidean metric), he would measure exactly the same length as you would (using the round metric). But the person outside the earth is allowed to travel along a larger class of paths than you are (e.g., the path through the earth) because his ambient manifold (all of $\mathbb{R}^3$) is larger than yours (the sphere/surface of the earth, which is a submanifold of $\mathbb{R}^3$). That's why in this case he can find a shorter path than you can.
To address your second question, the person on the earth does not live on an $xy$-plane, she lives on a sphere. I think this is an important point. The plane and the sphere are not only geometrically distinct (i.e., not isometric), but they are also topologically distinct (not homeomorphic). You seem to be suggesting that there are coordinates $(x,y)$ on the sphere in which the metric is $dx^2 + dy^2$, but this is incorrect. First of all, note that such coordinates could only possibly be defined locally, on some proper subset of the sphere, since the sphere is not homeomorphic to any subset of the plane. More fundamentally, even if you choose local coordinates $(x,y)$ on a subset of the sphere, the metric could never be the flat metric (in those or any other coordinates), since the sphere is not flat, which can be seen, for example, by computing that its Gauss curvature is nonzero.
Now, you could also, after choosing local coordinates $(x,y)$ on the sphere (in some neighborhood of Toronto), consider the flat metric $dx^2 + dy^2$. This would be a valid Riemannian metric in the coordinate neighborhood, but it would be completely irrelevant to this situation, because it would have nothing to do with the actual ("round") geometry of the earth, which, again, is induced from the flat metric on the ambient $\mathbb{R}^3$. It would be possible to choose the coordinates in such a way such that the distance between Toronto and Brazil is the same in either metric, but chances are that the distances between most points not on that great circle would be different in the two metrics.
Let me say a word about distances. It's important not to conflate the Riemannian metric, call it $g$, on a manifold $M$ and the metric, call it $d$, that makes $M$ a metric space. $g$ computes the inner product between pairs of tangent vectors, whereas $d$ computes the distance between pairs of points. Of course, $g$ detemines $d$, since the distance $d(p, q)$ is the infimum of lengths of curves joining $p$ and $q$ (and the length of a curve is computed by integrating the $g$-length of its velocity vector). The question of whether $d$ determines $g$ (which I think is related to your last question) is a harder one, and I'm not sure what can be said.