I have a question regarding how the gradient is related to the tangent and normal of curves / surfaces in 2 and 3 dimensions.
From what I've gathered, the gradient of a surface in $\mathbb{R}^3$ is normal to the surface at a specific point.
To illustrate, take the surface $x^2 + y^2 - z=0$. Calculating the gradient of $f(x,y,z) = x^2+y^2-z$ gives us $\nabla f(x, y, z) = (2x, 2y, -1)$. Evaluating the gradient at $(2, 1, 5)$ gives us a normal to the surface $f=0$ at that point: $\nabla f(2, 1, 5) = (4, 2, -1)$.
Now, for the gradient of a curve in $\mathbb R^2$ this does not seem to the case. Take the curve $f(x, y) = y - x^2$. Calculating the gradient gives us $\nabla f(x, y) = (-2x, 1)$. If we evaluate the gradient at $(1, 1)$ we get $\nabla f(1, 1) = (-2, 1)$. Converting this into a line we get $y = 2x$. This gives us a tangent to the curve $f=0$ (that is, $y = x^2$) at $(1, 1)$.
How does these two cases relate and why is the gradient to a curve in $R^2$ not the normal of the curve?
The mistake is when you "convert into a line". I don't know exactly what you did. However, if you just focus on the vector $(-2,1)$ on the $xy$ plane (draw it!), then this vector is clearly normal to the parabola in the point $x=1$. The line having direction $(-2,1)$ and passing by the point $(1,1)$ is $ y=3/2-x/2$, not $y=2x$. This line is not even passing by $(1,1)$!