Hartshorne Excercise II.2.12 is the following:
Gluing Lemma: Let $\{X_i\}$ be a (possibly infinite) family of schemes. For each $i\neq j,$ suppose we are given an open subset $U_{ij}\subseteq X_i,$ and let it have the induced scheme structure. Suppose also that we are given for each $i\neq j$ an isomorphism of schemes $\phi_{ij} : U_{ij}\to U_{ji}$ such that
- for each $i,j,$ $\phi_{ij} = \phi_{ji}^{-1},$ and
- for each $i,j,k,$ $\phi_{ij}(U_{ij}\cap U_{ik}) = U_{ji}\cap U_{jk},$ and $\phi_{ik} = \phi_{jk}\circ\phi_{ij}$ on $U_{ij}\cap U_{ik}.$
Then there exists a scheme $X,$ together with morphisms $\psi_i : X_i\to X$ for each $i,$ such that
- $\psi_i$ is an isomorphism of $X_i$ onto an open subscheme of $X,$
- the $\psi_i(X_i)$ cover $X,$
- $\psi_i(U_{ij}) = \psi_i(X_i)\cap\psi_j(X_j),$ and
- $\psi_i = \psi_j\circ\phi_{ij}$ on $U_{ij}.$
We say that $X$ is obtained by gluing the schemes $X_i$ along the isomorphisms $\phi_{ij}.$
The classic example of this procedure is the construction of projective space from affine spaces. It seems you're familiar with this example already, but understanding it deeply will reveal how gluing should work in similar situations (including yours), we let's review it. The projective line $\Bbb{P}^1$ is often described as gluing two copies of $\Bbb{A}^1$ via the map $x\mapsto \frac{1}{y},$ where $x$ is the coordinate of the first $\Bbb{A}^1$ and $y$ is the coordinate of the second $\Bbb{A}^1.$ Fix an algebraically closed field $k$ throughout this example (for simplicity).
In terms of the gluing lemma, what this is really saying is that we have the schemes $X_1 = \operatorname{Spec}k[x]$ and $X_2 = \operatorname{Spec}k[y],$ together with $U_{12} = D(x) = \operatorname{Spec}k[x,x^{-1}]\subseteq X_1$ and $U_{21} = D(y) = \operatorname{Spec}k[y,y^{-1}]\subseteq X_2$ and the morphisms
$$
\phi_{12} : \operatorname{Spec}k[x,x^{-1}]\to\operatorname{Spec}k[y,y^{-1}]
$$
and
$$
\phi_{21} : \operatorname{Spec}k[y,y^{-1}]\to\operatorname{Spec}k[x,x^{-1}]
$$
induced by the isomorphisms
\begin{align*}
f_{12} : k[y,y^{-1}]&\to k[x,x^{-1}]\\
y&\mapsto 1/x
\end{align*}
and
\begin{align*}
f_{21} : k[x,x^{-1}]&\to k[y,y^{-1}]\\
x&\mapsto 1/y,
\end{align*}
respectively.
If $\mathfrak{m}$ is a prime ideal of $k[x]$ (respectively $k[y]$), denote by $[\mathfrak{m}]$ the associated point in $X_1$ (respectively $X_2$). If we denote the embedding of $X_1\subseteq\Bbb{P}^1$ on closed points by $[(ax - b)]\mapsto [b : a],$ the equality $(ax - b) = (x - b/a)$ forces us to have $[b : a] = [b/a : 1]$ for nonzero $a$. More generally, this means that the homogeneous coordinates $[b:a]$ on $\Bbb{P}^1$ must satisfy the familiar rule $[bc : ac] = [b : a]$ for $c\in k^\times.$
Moreover, if $\mathfrak{m} = (ax - b)\subseteq k[x]$ ($a,b\neq 0$) then $\phi_{12}([\mathfrak{m}]) = [(by - a)].$ This means that the embedding of $X_2$ into $\Bbb{P}^1$ must be given by $[(by - a)]\mapsto [b : a]$ for $a,b\in k^\times.$ We may extend this compatibly to all of $X_2$ by saying that $[(by - a)]\mapsto [b : a]$ for all $a\in k, b\in k^\times.$
Notice that all this implies that if we simply view a point $[(ax - b)]\in X_1$ as the coordinate $b/a,$ and if $b\neq 0,$ then $b/a$ is identified with the coordinate $a/b$ from $X_2$ inside $\Bbb{P}^1$ via $[b/a : 1] = [1 : a/b].$
There's a nice picture and more descriptions of this in Vakil's book as example 4.4.6, and you can probably find it in your favorite introductory text on algebraic geometry as well.
So what are we doing in your example? Well, we can translate the above relatively directly. Here we have
$$
C_0 = \operatorname{Spec}k[x,y]/(y^2 - (x^4 - 7)),
$$
and
$$
C_1 = \operatorname{Spec}k[u,v]/(v^2 - (1 - 7u^4)).
$$
Let $U_{01} = D(x) = \operatorname{Spec}k[x^{\pm1},y]/(y^2 - (x^4 - 7))\subseteq C_0$ and $U_{10} = D(u) = \operatorname{Spec}k[u^{\pm1},v]/(v^2 - (1 - 7u^4))\subseteq C_1.$
Next consider the ring maps
\begin{align*}
f_{01} : k[u^{\pm1},v]/(v^2 - (1 - 7u^4))&\to k[x^{\pm1},y]/(y^2 - (x^4 - 7))\\
u&\mapsto 1/x\\
v&\mapsto y/x^2,
\end{align*}
and
\begin{align*}
f_{10} : k[x^{\pm1},y]/(y^2 - (x^4 - 7))&\to k[u^{\pm1},v]/(v^2 - (1 - 7u^4))\\
x&\mapsto 1/u\\
y&\mapsto v/u^2.
\end{align*}
These induce morphisms $\phi_{01} : U_{01}\to U_{10}$ and $\phi_{10} : U_{10}\to U_{01},$ respectively.
You should then check that $\phi_{01}$ and $\phi_{10}$ are in fact isomorphisms, so that the gluing lemma may be applied. Assuming that we have checked this, we obtain a new scheme $C$ which is covered by the two curves $C_0$ and $C_1$ such that if we consider a closed point $(a,b)\in C_0(k)$ with $a\neq 0,$ the corresponding closed point in $C(k)$ is identified with the point corresponding to $(1/a,b/a^2)\in C_1(k).$
However, notice that the scheme $C$ provided by the gluing lemma is not guaranteed to sit inside $\Bbb{P}^2$ -- it is only guaranteed to exist as an abstract scheme. And indeed, the identifications between the $U_{01}$ and $U_{10}$ are not the same as the identifications that one would obtain when trying to glue multiple copies of $\Bbb{A}^2$ to get $\Bbb{P}^2,$ so we should not expect this curve $C$ to live inside $\Bbb{P}^2$ in the "usual" way via this construction. This should make sense intuitively as well: if we embed $C_0$ inside $\Bbb{P}^2$ in this usual way and take the closure we obtain a singularity, so there's something problematic about that choice of embedding.
The gluing described in your question instead sits naturally inside the weighted projective space $\Bbb{P}(1,2,1).$ This space is defined as the quotient of $\Bbb{A}^3\setminus\{0\}$ by the $\Bbb{G}_m$ action given by $\lambda\cdot (a_0,a_1,a_2) := (\lambda a_0, \lambda^2 a_1, \lambda a_2).$ The homogeneous coordinates in this space satisfy $[x : y : z] = [\lambda^1 x : \lambda^2 y : \lambda^1 z]$ for $\lambda\in k^\times,$ instead of the familiar $[x : y : z] = [\lambda x : \lambda y : \lambda z].$ You can read more about the construction and properties of weighted projective space here or here.
The embedding of $C_0\to\Bbb{P}(1,2,1)$ sends $(x,y)\mapsto [x : y : 1],$ and the embedding of $C_1\to\Bbb{P}(1,2,1)$ sends $(u,v)\mapsto [1 : v : u].$ Then we see that if $a\neq 0,$ we have
$$
[a : b : 1] = [a^{-1}a : a^{-2} b : a^{-1} 1] = [1 : b/a^2 : 1/a],
$$
which is precisely what we would expect, given that the identification between $U_{01}\to U_{10}$ sends $(a,b)\mapsto (1/a, b/a^2).$
