Here is a reasonably geometric way to move $\Delta_X$ to some other (non-effective) divisor.
Strategy. For $\mathbf P^1$, we know how to do this (see below for an alternative method). In general, take a sufficiently nice finite map $f \colon X \to \mathbf P^1$, and consider $g = f \times f \colon X \times X \to \mathbf P^1 \times \mathbf P^1$. Then $\Delta_X$ is a component of $D = g^{-1}(\Delta_{\mathbf P^1})$, and we know how to move $D$. If $D' = D - \Delta_X$ is the 'residual divisor', it remains to compute $\Delta_X \cdot D$ and $\Delta_X \cdot D'$.
Spoiler: if $d = \deg f$ and $R$ is the ramification divisor [Hart77, §IV.2], then $\Delta_X \cdot D = 2d$ and $\Delta_X \cdot D' = \deg R$, so
$$\Delta_X^2 = \Delta_X \cdot D - \Delta_X \cdot D' = 2d - \deg R,$$
which gives the result by Riemann–Hurwitz!
Example. For $\mathbf P^1$, there are many methods to do this; see for instance Donu Arapura's comment. Another method is as follows: any non-identity matrix $A \in \operatorname{PGL}_2(k)$ gives a graph $\Gamma_A \subseteq \mathbf P^1 \times \mathbf P^1$ that is linearly equivalent to $\Delta_{\mathbf P^1}$. The intersections of $\Delta_{\mathbf P^1}$ with $\Gamma_A$ are given by the eigenvectors of $A$, so there are $2$ when counted with multiplicity. If you want it even more explicitly, you can take $A$ to be diagonal, so the intersection points are $([0:1],[0:1])$ and $([1:0],[1:0])$.
Remark. In the general case, we pick a suitable finite morphism $f \colon X \to \mathbf P^1$. For instance, we probably want $f$ to be separable (which is always possible; see e.g. this post). And it might even be desirable to have $f$ be tamely ramified, which is possible if $\operatorname{char} k \neq 2$ or $k \subseteq \bar{\mathbf F}_2$, but not in general [KLW23]. In fact, for $\operatorname{char} k \neq 2$, a sufficiently general $f$ has ramification index $2$ at each ramification point in $x$.
Let $g = f\times f$ and $D = g^{-1}(\Delta_{\mathbf P^1})$ as before, and set $D' = D - \Delta_X$. Note that the support of $D'$ is the closure of the locus $\{(x,y) \in X^2\ |\ x \neq y \text{ and } f(x) = f(y)\}$.
The first claim is that $D$ is geometrically reduced. After immediate passage to $\bar k$, by [Tag 0344] we have to show that $D$ is (R0) and (S1). The second is automatic as $D$ is a divisor in a smooth variety [Tag 00SB]. To see that $D$ is generically smooth, take an open $U \subseteq X$ such that $f|_U$ is étale (which exists as $f$ is separable). Then $D \cap (U \times U) = U \times_{\mathbf P^1} U$ is smooth since it is étale over $\mathbf P^1$. Since $D$ contains no horizontal or vertical divisors, $D \cap (U \times U)$ is dense in $D$.
Next we compute $\Delta_X \cdot D$ by moving $D$ to the linearly equivalent divisor $E = g^{-1}(\Gamma_A)$. If $d = \deg f$, then the projection formula [Hart77, Appendix A, §1, A4] gives
$$\Delta_X \cdot D = \Delta_X \cdot E = (g_* \Delta_X) \cdot \Gamma_A = (d \Delta_{\mathbf P^1}) \cdot \Gamma_A = 2d.$$
In this case, the projection formula for divisors can actually be deduced from that for coherent sheaves [Hart77, Exc. II.5.1(d)], using the formula $D_1 \cdot D_2 = \operatorname{length}(\mathcal O_{D_1} \otimes \mathcal O_{D_2})$ and the fact that pushforward along the degree $d$ locally free morphism $\Delta_X \to \Delta_{\mathbf P^1}$ multiplies length by $d$. Or dispose of the projection formula altogether by using $D_1 \cdot D_2 = \deg(\mathcal O_S(D_2)|_{D_1})$, which is well-suited to the case where $D_2$ is pulled back from somewhere.
To finish the computation, it remains to compute $\Delta_X \cdot D'$. Clearly the intersection points of $\Delta_X$ with $D'$ are exactly the ramification points of $f$. But in fact, $\Delta_X \cdot D' = \deg R$ is the degree of the ramification divisor, defined as in [Hart77, §IV.2]. For instance, if $f$ has ramification index $2$ everywhere and $\operatorname{char} k \neq 2$, this says that $D'$ meets $D$ transversely, which is easily checked in local coordinates.
For an actual proof (in general) of $\Delta_X \cdot D' = \deg R$, we use the short exact sequence
$$0 \to \mathcal O_{X \times X}(-\Delta_X)\big|_{D'} \to \mathcal O_D \to \mathcal O_{\Delta_X} \to 0$$
obtained locally from
$$0 \to \mathcal O_V/(g) \stackrel f\to \mathcal O_V/(fg) \to \mathcal O_V/(f) \to 0$$
on an open subset $V \subseteq X \times X$ on which $\Delta_X$ is cut out by $f$ and $D'$ by $g$ (exactness on the left uses that $X \times X$ is locally factorial). We conclude that the ideal $\mathscr I \subseteq \mathcal O_D$ cutting out $\Delta_X$ is isomorphic to $\mathcal O_{X \times X}(-\Delta_X)|_{D'}$, so
$$\deg R = \operatorname{length}\left(\mathscr I/\mathscr I^2\right) = \operatorname{length}\left(\mathcal O_{X \times X}(-\Delta_X)\big|_{D' \cap \Delta_X}\right),$$
which equals $\Delta_X \cdot D' = \operatorname{length}(\mathcal O_{D' \cap \Delta_X})$ since line bundles on zero-dimensional schemes are trivial. $\square$
References.
[Hart77] R. Hartshorne, Algebraic geometry. Graduate Texts in Mathematics 52. Springer-Verlag, 1977. ZBL0367.14001.
[KLW23] K. S. Kedlaya, D. Litt, J. Witaszek, Tamely ramified morphisms of curves and Belyi’s theorem in positive characteristic. Int. Math. Res. Not. 2023.4 (2023), p. 2803-2833. ZBL1509.14048.
