A linear transformation is one where
$T(\alpha x) = \alpha T(x)$
$T(x + y ) = T(x) + T(y)$
That's all the is to it.
$T((x, y)) = (2x + y, y, 0)$
Is linear because $T(c(x,y)) = T((cx,cy)) = (2cx + cy, cy, 0) = c(2x + y, cy, 0) = cT((x,y))$
and $T((x,y) + (w,z)) = T((x+w, y + z)) = (2x + 2w + y + z, y + z, 0) = (2x +y,y,0) + (2w + z, z,0) = T((x,y)) + T((w,z))$
But $T((x,y)) = (2x + y, xy, 0)$ is not because
$T(c(x,y)) = (2cx + cy, c^2xy, 0) \ne (2cx + cy, cxy, 0) = cT((x,y))$.
Also $T((x,y) + (w,z)) \ne T((x,y)) + T((w,z))$
A trick to notice is if the transformation involves adding a scalar other than 0 it will not be linear. ($f(x) = x + c => f(2x) = 2x + c; c \ne 2c: f(x + y) = x + y + c \ne x + c + y + c$)
If the transformation involves multiplying two terms together or taking a power of a term it will not be linear. ($f(x) = x^2 => f(2x) = (2x)^2 \ne 2(x^2): f(x + y) = (x +y)^2 \ne x^2 + y^2$)
But if the transformation is only multipling terms by scalars and adding terms it is linear.
$f((x,y)) = \sum (c_ix + d_iy) \implies f(a(x,y) + e(w,z)) = \sum (c_iax + d_iay + ec_iw + d_iez) = a\sum(c_ix + d_iy) + e\sum(c_ix + d_iy) = af((x,y)) + ef((w,z))$
Intuitive. If you travel along a line the distance you change vertically is proportional to the distance you change horizontally. Lines are consistant. If you double the input, you double the output. If you add two inputs together you get the two outputs added together.
If you travel anything that isn't a line that isn't true any more. That's why they call them "linear".
BUT there is a caveat. In high school algebra lines could have y-intercepts which throw these proportions off. It's only the "steepness" that is linear. In linear algebra they call those type of functions "affine". They are sort of "linear" but with a constant "slide displacement to them".