You are missing the crucial independence and zero-mean conditions, without which the Marcinkiewicz–Zygmund inequalities will not hold in general.
These inequalities are actually as follows: for each real $p\ge1$, some positive real $A_p$ and $B_p$ and all $n$-tuples $X:=(X_1,\ldots,X_n)$ of independent zero-mean random variables (r.v.'s) \begin{equation*} A_p E\|X\|_2^p\le E|S_X|^p\le B_p E\|X\|_2^p, \tag{1}\label{1} \end{equation*} where $S_X:=\sum_1^n X_i$ and $\|X\|_2:=(\sum_1^n X_i^2)^{1/2}$.
Consider first the case when the $X_i$'s are symmetric(ally) distributed. Then, conditioning on the $|X_i|$'s, we see that the best constants $A_p$ and $B_p$ in \eqref{1} are the same as the best constants $A_p$ and $B_p$ in the case when the $X_i$'s are real scalar multiples of Rademacher r.v.'s (see details on this at the end of this answer). In the latter, scalar-multiples-of-Rademacher case, the best constants $A_p$ and $B_p$ were obtained by Haagerup; let us denote these symmetric-case constants by $A_{p,S}$ and $B_{p,S}$, respectively.
Given now arbitrary independent zero-mean r.v.'s $X_1,\ldots,X_n$, we can symmetrize them by considering the $n$-tuple $X-X'$ of symmetric independent r.v.'s $X_i-X'_i$, where $X':=(X'_1,\ldots,X'_n)$ is an independent copy of $X=(X_1,\ldots,X_n)$. Then, by Jensen's inequality, \begin{equation*} E\|X\|_2^p\le E\|X-X'\|_2^p\le2^p E\|X\|_2^p \end{equation*} and \begin{equation*} E|S_X|^p\le E|S_X-S_{X'}|^p\le2^p E|S_X|^p. \end{equation*} So, \begin{equation*} E|S_X|^p\le E|S_X-S_{X'}|^p=E|S_{X-X'}|^p \\ \le B_{p,S}E\|X-X'\|_2^p \le2^p B_{p,S}E\|X\|_2^p \end{equation*} and \begin{equation*} 2^p E|S_X|^p\ge E|S_X-S_{X'}|^p =E|S_{X-X'}|^p \\ \ge A_{p,S}E\|X-X'\|_2^p \ge A_{p,S}E\|X\|_2^p. \end{equation*} So, \eqref{1} holds for each real $p\ge1$ and all $n$-tuples $X:=(X_1,\ldots,X_n)$ of independent zero-mean r.v.'s with \begin{equation*} A_p=A_p^*:=2^{-p}A_{p,S}\quad\text{and}\quad B_p=B_p^*:=2^p B_{p,S}. \end{equation*} In particular, letting here $p=1$, we get \begin{equation*} A_1^*:=2^{-3/2}\quad\text{and}\quad B_1^*:=2. \end{equation*} However, again by Jensen's inequality, for all $p\in[1,2]$ and, in particular, for $p=1$, \begin{equation*} E|S_X|^p\le(ES_X^2)^{p/2}=E\|X\|_2^p, \end{equation*} so that the second inequality in \eqref{1} actually holds with $B_p=1$.
Details on the conditioning: Here the conditioning on the $|X_i|$'s can be done simply as follows: Note that the joint distribution of the symmetric r.v.'s $X_1,\ldots,X_n$ is the same as the joint distribution of the r.v.'s $R_1|X_1|,\ldots,R_n|X_n|$, where $R_1,\ldots,R_n$ are independent Rademacher r.v.'s (with $P(R_i=\pm1)=1/2$) that are also independent of $X_1,\ldots,X_n$. So, for $Z:=(R_1|X_1|,\ldots,R_n|X_n|)$, first we average $|S_Z|^p$ and $\|Z\|_2^p$ with respect to the $R_i$'s with values of the $|X_i|$'s fixed and thus get
\begin{equation*}
A_{p,S} E\big(\|Z\|_2^p\big|\,|X|\big)
\le \big(E|S_Z|^p\big|\,|X|\big)
\le B_{p,S} E\big(\|Z\|_2^p\big|\,|X|\big), \tag{2}\label{2}
\end{equation*}
where $|X|:=(|X_1|,\ldots,|X_n|)$, and
$A_{p,S}$ and $B_{p,S}$ are Haagerup's constant factors, as before.
Next, we average the terms in \eqref{2} with respect to the $|X_i|$'s, to remove the conditioning on $|X|$ and thus get
\begin{equation*}
A_{p,S} E\|Z\|_2^p
\le E|S_Z|^p
\le B_{p,S} E\|Z\|_2^p.
\end{equation*}
Finally, we recall that $Z$ has the same distribution as $X$.