Shafarevich says the following:
Using Lüroth's theorem, we see that if $X$ is a rational curve, then $k(X)$ is isomorphic to the field of rational functions $k(t)$.
This is equivalent to saying that we can somehow prove that $t\in k(X)$. How is this true if the curve is $(t,t^2+1)$?
Here presumably $X$ is a rational curve in some affine (or projective, makes no difference) curve, $X \subset \Bbb{A}^n$. So the points $P$ of $X$ are gotten from a recipe like $P=(x_1,x_2,\ldots,x_n)$, where $x_i=g_i(t)$ for some rational functions $g_1(t),g_2(t),\ldots,g_n(t)\in k(t)$. The function field of $X$ is thus $$ k(X)=k(g_1(t),g_2(t),g_3(t),\ldots,g_n(t)). $$ Because the functions $g_i(t)\in k(t)$, it follows that $k(X)\subseteq k(t)$. Because $X$ is a curve, the field $k(X)$ contains elements that are $\notin k$. Therefore Lüroth's theorem kicks in, and we can conclude that $k(X)=k(u)$ for some element $u\in k(t)$. For all $i=1,2,\ldots,n$ we have $g_i(t)\in k(X)$, so this means that $g_i(t)=h_i(u)$ for some rational function $h_i$.
In the example case $$ X=\{(t,t^2+1)\in\Bbb{A}^2\mid t\in k\} $$ there is no problem, as $$ k(X)=k(t,t^2+1)=k(t). $$ Lüroth's theorem comes to the fore, when we use more complicated parametrizations. For example, it is not immediately clear what is the natural parameter $u$ for a curve like $$ X=\{(\frac{t+t^2}{1+t^5+t^6+t^9+t^{10}},\frac{1+t^3+t^4+t^5+t^6}{t+t^8})\mid t\in k\}. $$