The complex equation $w - z = 0$, $z$, $w \in \mathbb{C}$, represents a complex curve (also called $1$-dimensional complex manifold). This complex curve corresponds to the complex plane $\mathbb{C}$ which is equivalent to the two dimensional real plane, $\mathbb{R^2}$.
If we compactify the complex plane, $\mathbb{C}$, then we obtain the closed complex plane, which is in one to one correspondence to the Riemann sphere .
My doubts are:
How does the complex curve $w - z = 0$ correspond to the complex plane?
How is the closed complex plane in one to one correspondence to the Riemann sphere?
$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Cpx}{\mathbf{C}}\newcommand{\Proj}{\mathbf{P}}$"The complex equation $w - z = 0$, $z$, $w \in \Cpx$..." implicitly views $w - z = 0$ as a constraint in $\Cpx^{2}$. The solution set is the set of pairs $(z, z)$ with $z$ complex, just as it would be in the more familiar setting of a real equation describing a locus in the real plane.
(Incidentally, $\Cpx$ is not equivalent to the real plane $\Reals^{2}$ in the context of your question; the copy of $\Cpx$ in your question is a complex line in $\Cpx^{2}$, while the real subspace $\Reals^{2}$ is not a complex subspace of $\Cpx^{2}$.
One way to compactify $\Cpx$ (and, it turns out, the only way to compactify to get a holomorphic curve) is to "add a single point at infinity". Intrinsically, the usual construction is to "glue" two copies of $\Cpx$, identifying each non-zero complex number $z$ in one copy with its reciprocal $w = 1/z$ in the other copy.
To see this construction gives a sphere, you can map $$ \{(r + is, t) \in \Cpx \times \Reals : r^{2} + s^{2} + t^{2} = 1\} \tag{1} $$ to $\Cpx$ by stereographic projection $\Pi_{N}$ from the north pole $N = (0, 0, 1)$, or by the complex conjugate of stereographic projection $\Pi_{S}$ from the south pole $S = (0, 0, -1)$. Writing $z = x + iy$ and $w = u + iv$, these mappings and their inverses are \begin{align*} z &= \Pi_{N}(r + is, t) = \frac{r + is}{1 - t}, & \Pi_{N}^{-1}(z) &= \frac{\bigl(2(x + iy), -(x^{2} + y^{2}) + 1\bigr)}{x^{2} + y^{2} + 1}; \\ w &= \overline{\Pi}_{S}(r + is, t) = \frac{r - is}{1 + t}, & (\overline{\Pi}_{S})^{-1}(w) &= \frac{\bigl(2(u - iv), -(u^{2} + v^{2}) + 1\bigr)}{u^{2} + v^{2} + 1}. \end{align*} Algebra gives $$ \Pi_{N} \circ (\overline{\Pi}_{S})^{-1}(w) = \frac{u - iv}{u^{2} + v^{2}} = \frac{1}{w}, $$ which shows the sphere (1) is in bijective correspondence with the "glued" complex lines.