I don't understand how (1.2.3) is equal to (1.2.2). I'm happy with everything up to (1.2.2) but I don't understand the explanation and step to (1.2.3) at all.
$\bar{u}^1,\bar{u}^2$ were not defined anywhere before this section. The Einstein summation notation is being used here.
I understand this isn't much to go on but I appreciate any alternative explanation you might be able to offer.
This is standard material when discussing surfaces (or manifolds in general), and it boils down to the chain rule.
For example, you could use standard cartesian coordinates $u^1=x$, $u^2=y$ on the plane and take the parametrization $$X(u^1,u^2) = (u^1,u^2),$$ or you could use polar coordinates (on a large open subset of the plane) $\bar u^1=r$, $\bar u^2=\theta$, and take the parametrization $$\bar X(\bar u^1,\bar u^2) = (u^1\cos u^2, u^1\sin u^2).$$ Then $X_1 = X_{u^1} = (1,0)$, $X_2 = X_{u^2} = (0,1)$ give the standard basis for the tangent plane at any point. Or you can use $\bar X_1 = \bar X_{\bar u^1} = (\cos u^2, \sin u^2)$ and $\bar X_2 = \bar X_{\bar u^2} = (-u^1\sin u^2,u^1\cos u^2)$ as the polar coordinate basis (written $\hat r$, $r\hat\theta$ in physics texts).
If you take a general tangent vector, you can write it as a linear combination of each set of basis vectors. The chain rule tells you how those linear combinations are related. Say a vector is written $\sum\xi^i X_i = \sum \bar\xi^j\bar X_j$. By the standard multivariable chain rule, we have $$\bar X_{\bar u^j} = \sum \frac{\partial u^i}{\partial \bar u^j} X_{u^i}.$$ Thus, $$\sum\bar\xi^j\bar X_j = \sum \bar\xi^j \bar X_{\bar u^j} = \sum \bar\xi^j \frac{\partial u^i}{\partial \bar u^j} X_{u^i}=\sum \xi^i X_{u^i},$$ which tells us that $$\xi^i = \sum \bar\xi^j \frac{\partial u^i}{\partial \bar u^j}.$$ Inversely (here you of course use invertibility of the Jacobian), $$\bar\xi^j = \sum \xi^i \frac{\partial\bar u^j}{\partial u^i}.$$