Given a perfectoid field $K$, for example $\widehat{\mathbf{Q}_p(\zeta_{p^{\infty}})}, \widehat{\mathbf{Q}_p(p^{1/p^{\infty}})}$ or $\widehat{\mathbf{Q}_p(\zeta_{p^{\infty}}, p^{1/p^{\infty}})}$. According to Bhatt's lecture note ( If I understand correct, see for example section 3.2), seems we have to first define the tilting of the ring of integer $K^{\circ\flat}$, then define $K^{\flat}$ to be the fraction field of the former, or more precisely invert certain element $t$ (see for example Prop 3.2.4). While on the other hand, notice there is a more "direct?" way to define the "tilting" directly as $\varprojlim_{\phi}K$ (If I am correct, this is exactly the $R(E)$ defined in the original paper of Wintenberger :see section 4.1)
Remark: Sure, you can not see directly what is the addition: in priori $\varprojlim_{\phi}K$ is only a multiplicative monoid, while the fact that there is a bijection $\varprojlim_{\phi}O_K/p\simeq \varprojlim_{\phi} O_K$ help us to "transfer" the addition structure from the former to the latter (via this bijection). But as in the original paper of Wintenberger section 4.1, you can just formally define that addition: $(x_n)+(y_n):=(z_n)$ with $z_n:=\lim_m(x_{n+m}+y_{n+m})^{p^m}$, to make it a characteristic $p$ field.
Question: Do these two definitions coincide? More precisely, do we have $Frac(\varprojlim_{\phi}O_K) \simeq \varprojlim_{\phi}K$?
Remark: There is an injective map (from left to right): it suffices to map $\frac{1}{y}$ for any $y=(y_n)_{n\in\mathbf{N}}\in (O_K)^{\flat}$, to $(\frac{1}{y_n})_{n\in \mathbf{N}}\in \varprojlim_{\phi}K$. While for the surjectivity it is a little subtle: take any $z=(z_n)_n\in \varprojlim_{\phi}K$, we can write $z_0=\frac{x_0}{y_0}$, and in our case we can construct $x=(x_n)_n$ and $y=(y_n)_n\in (O_K)^{\flat}$ and expect $x/y$ is a preimage. But then you notice, first position is by construction $\frac{x_0}{y_0}=z_0$, good, but the second position is $\frac{x_1}{y_1}$ which is indeed a $p$-th root of $z_0$, but not necessarily the $z_1$ (Notice if we were working in characteristic $p$, then they must coincide $z_1=\frac{x_1}{y_1}$ as "$x^p=y^p \Longleftrightarrow x=y$ in char $p$"). Is this indeed THE problem, or actually we can fix that?
I summarize my comment as an answer. Basically this is exactly Lemma 3.4 (iii) in the paper Scholze's original paper.
We first construct $\varprojlim\limits_{x\to x^p}O_K/p\to O_K, x=(\overline{x_n})_n\mapsto x^{\sharp}:=\lim_\limits{n\to \infty}x_n^{p^n}$ where $x_n$ is any lift from $O_K/p$ to $O_K$. Actually this limit exists and is independent of choice of lifts.
Now we construct $\varprojlim\limits_{x\to x^p}O_K/p\to \varprojlim\limits_{x\to x^p}O_K$, we construct the map $x\mapsto (x^{\sharp}, (x^{\sharp})^{1/p},(x^{\sharp})^{1/p^2}, \cdots)$ and because $x\mapsto x^{\sharp}$ is multiplicative, this extends to $(-)^{\sharp}: Frac(\varprojlim\limits_{x\to x^p}O_K/p)\to \varprojlim\limits_{x\to x^p}K$. We call the componentwise mod $p$ from right to left $\mathsf{Proj}$, and one checks this two map are inverse to each other.
Indeed, let's check the less obvious direction (corresponds to the surjectivity): $(-)^{\sharp}\circ \mathsf{Proj}=\mathsf{Id}$. Take $x=(x_0, x_1, x_2, \cdots)$ from the RHS, then we check $(-)^{\sharp}\circ \mathsf{Proj}(x)=x$. Let's compute the $m$-th position: by construction it is
(we use $\widetilde{x_t}$ to denote an arbitrary lift of $\overline{x_t}$ to $O_K$): the last equality comes from the fact we are in a $p$-adic complete ring plus the observation ($x\equiv y$ mod $p) \Rightarrow (x^{p^n}\equiv y^{p^n}$ mod $p^n$).
Summary："In characteristic $0$, mod $p$ indeed lose information, but if you know mod $p$ of all its $p$-power roots, then you don't lose information (in a $p$-adic complete ring)"