$\def\frq{\mathfrak{q}}$Exercise 1. If $S$ is a reduced ring of characteristic $p$ and $s\in S$, then $x^p-s\in S[x]$ either has no roots or has a unique root of multiplicity $p$.
Thus, if $x^p-s$ has a root in $S$, the symbol $\sqrt[p]{s}$ denotes a unique element of $S$.
In the following, $R$ is a reduced ring of prime characteristic $p$.
We will define $R^{1/p}$ via a universal property. Before, let us introduce some definitions: We say that an $R$-algebra $S$ is closed under $p$-th $R$-roots if for each $r\in R$ there is $s\in S$ such that $s^p=g(r)$, where $g:R\to S$ is the structure morphism, i.e., $S$ has all $p$-th roots of $g(R)$. Consider the full subcategory $\mathsf{C}\subset R\text{-}\mathsf{Alg}$ of the reduced $R$-algebras that are closed under $p$-th $R$-roots. We define $R^{1/p}$ to be an initial object in $\mathsf{C}$.
We just need to show $R^{1/p}$ exists. We claim it is the Frobenius homomorphism $F:R\to R$. Let $g:R\to S$ be an object in $\mathsf{C}$. We want to factor $g$ through $F$. Define $\overline{g}:R\to S$ to be the map that sends $r\in R$ to $\sqrt[p]{g(r)}\in S$. By uniqueness of $p$-th roots and the fact that $(-)^p:S\to S$ is a ring homo, it is easy to verify that $\overline{g}$ is a ring homomorphism. By construction, $\overline{g}\circ F=g$. We now see uniqueness of $\overline{g}$: if $\tilde{g}$ were another such factorization, then $\overline{g}(r)^p=\overline{g}(r^p)=g(r)=\tilde{g}(r^p)=\tilde{g}(r)^p$. Since $S$ is reduced, $(-)^p:S\to S$ is injective, so $\overline{g}(r)=\tilde{g}(r)$.
Lemma 2. An object $g:R\to S\in\mathsf{C}$ is initial in $\mathsf{C}$ if and only if $g(R)=S^p$.
Proof. On the one hand, if $g$ is initial in $\mathsf{C}$, then $g\cong F$ so the claim is trivial. The proof of the converse is left as an exercise, since it is very similar to the proof that $F$ is initial. $\square$
Exercise 3. If $S$ is a reduced ring of characteristic $p$ and $s\in S$ is such that $x^p-s\in S[x]$ has no roots in $S$, then $S[x]/(x^p-s)$ is reduced (i.e., $(x^p-s)\subset S[x]$ is a radical ideal).
With the universal-property definition, we can write the Frobenius map $F:R\to R$ in a quite suggestive way.
Lemma 4. $F$ is isomorphic to the unique $R$-algebra map
$$
f:R\to
\frac{R[x_r\mid r\in R,\sqrt[p]{r}\not\in R]}
{(x_r^p-r\mid r\in R,\sqrt[p]{r}\not\in R)}=R'
$$
By “$\sqrt[p]{r}\not\in R$” we mean “there is no $p$-th root of $r$ in $R$.” The fact $F\cong f$ makes precise the idea that “$R^{1/p}$ is obtained by formally adding to $R$ a $p$-th root of each element that hadn't one.”
Proof. We need to verify two things: that $f$ is an object in $\mathsf{C}$ and that it is initial in $\mathsf{C}$. Suppose we knew $R'$ is reduced. Then it is easy to show that $f$ is an object of $\mathsf{C}$ which is initial (Lemma 2). Thus, we just need to verify $R'$ is reduced. While undertaking the search for a proof of the reducedness of $R'$, one might encounter the following difficulty: if $r,s\in R$ are distinct elements of $R$ with no $p$-th root in $R$, then the ideal $(x^p-r,y^p-s)$ might no be radical in $R[x,y]$. Intuitively, this is because adjoining one single $p$-th root to $R$ might automatically add other $p$-th roots: Given an element $r\in R$, it is easy to show by induction that $\sqrt[p]{r}\in R$ if and only if $\sqrt[p]{r+n}\in R$, for all $n\in\mathbb{Z}$. We can overcome this difficulty in the following way: we will use the result a filtered colimit of reduced rings is reduced (it follows from this). Let's write then $R'$ as a filtered colimit of reduced rings: For each $T\subset R$, write
$$
R_T=\frac{R[x_r\mid r\in T]}{(x_r^p-r\mid r\in T)}.
$$
Define a category $\mathsf{I}$, whose objects are finite subsets $T\subset R$ such that $R_T$ is reduced (thus $\sqrt[p]{r}\not\in R$ for all $r\in T$; note the converse might not hold, because of the phenomenon pointed out before). Given objects $T_1,T_2\in \mathsf{I}$, we observe that there is at most one morphism of $R$-algebras $R_{T_1}\to R_{T_2}$ (use Exercise 1). In such a case, we will say there is a unique morphism $T_1\to T_2$ in $\mathsf{I}$ (so $\mathsf{I}$ is a thin small category).
Define a diagram
\begin{align*}
D:\mathsf{I}&\to\mathsf{Ring}\\
T&\mapsto R_T
\end{align*}
where the action on morphisms sends $T_1\to T_2$ to the unique $R$-algebra map $R_{T_1}\to R_{T_2}$.
We first verify that $R'$ is the colimit of this diagram. It is clear that $R'$ is a cocone under $D$. It is a limiting cocone since given a cocone $\mathcal{A}=\{R_T\to A\}_{T\in\mathsf{I}}$ under $D$, in particular we have maps $\varphi:R=R_\varnothing\to A$ and $\psi_r:R_{\{r\}}\to A$, for each $r\in R$ such that $\sqrt[p]{r}\not\in R$ ($\{r\}$ is in $\mathsf{I}$ by Exercise 3). Thus, there is a canonical map $R'\to A$ that acts as $\varphi$ on coefficients and that sends the coset of $x_r$ in $R'$ to $\psi_r(x_r)\in A$. It is not difficult to see that $R'\to A$ factors $\mathcal{A}$. Uniqueness of the factorization $R'\to A$ comes from the fact that a ring morphism $R'\to B$ is fully determined by the composites with $R=R_\varnothing\to R'$ and with $R_{\{r\}}\to R'$, where $r\in R$ is such that $\sqrt[p]{r}\not\in R$ (i.e., $R'\to B$ is determined by the action on coefficients in $R$ and the action on the cosets of the $x_r$'s).
It is left to see that $\mathsf{I}$ is a filtered category. On the one hand, it is non-empty, for $\varnothing\in\mathsf{I}$. Since $\mathsf{I}$ is thin, we just need to show that given $T_1,T_2\in\mathsf{I}$, there is $T\in\mathsf{I}$ such that there are morphisms $T_1\to T\leftarrow T_2$ in $\mathsf{I}$. Note that there is an $R$-algebra morphism $T_1\to T_2$ in $\mathsf{I}$ if and only if all $r\in T_1$ have a $p$-th root in $R_{T_2}$. If there is an $R$-algebra morphism $T_1\to T_2$, we are done. Otherwise, pick some $r\in T_1$ without a $p$-th root in $R_{T_2}$. Then $R_{T_2\cup\{r\}}=R_{T_2}[x]/(x^p-r)$ is reduced (Exercise 3), i.e., $T_2\cup\{r\}\in\mathsf{I}$. We may replace $T_2$ by $T_2\cup\{r\}$ (for there is a morphism $T_2\to T_2\cup\{r\}$ in $\mathsf{I}$). Now $r$ has a $p$-th root in $R_{T_2}$. If at this moment there is $T_1\to T_2$, we are done; otherwise, we repeat the same process. Since $T_1$ is finite this process finds a morphism $T_1\to T_2$ in a finite number of steps, and we win. $\square$
There are two special cases in which $R^{1/p}$ has other presentations:
If $R$ is an integral domain, then we can consider its field of fractions $Q(R)$ and embed $R$ in an algebraic closure $R\subset\overline{Q(R)}$. Define $S=\{s\in\overline{Q(R)}\mid s^p\in R\}$. Then it is easy to verify that $S$ is a subring of $\overline{Q(R)}$. We have that $S$ is an $R$-algebra closed under $p$-th $R$-roots, and it is initial in $\mathsf{C}$ thanks to Lemma 2.
In the case when $R$ has finitely many minimal primes $\frq_1,\dots,\frq_t$ (this happens for instance if $R$ is Noetherian), we can give yet another construction for $R^{1/p}$. In this case, by 00EW(3) we are under the hypotheses of 02LX. Thus the total ring of fractions of $R$ is $Q(R)=R_{\frq_1}\times\cdots \times R_{\frq_t}$. By 00EU, each factor $R_{\frq_i}$ is a field. Taking algebraic closure in each factor, we have an inclusion $R\subset K:=\overline{R_{\frq_1}}\times\cdots \times \overline{R_{\frq_t}}$. Define $S=\{s\in K\mid s^p\in R\}$. Then it is easy to verify that $S$ is a subring of $K$. On the other hand, $S$ is reduced (for $K$, being a product of reduced rings, is reduced), and $R\subset S$, so that $S$ is an $R$-algebra closed under $p$-th roots. By Lemma 2, it is initial in $\mathsf{C}$, i.e., $S=R^{1/p}$.
