This is not true.

(2) $\not\Rightarrow$ (1)

Let $f=\mathrm{id}_X$, then $F_s$ is supported at a point, so no matter what, $F_s$ is not $S_1$. So, you just have to choose an $X$ and and $F$ that's $S_1$.

(1) $\not\Rightarrow$ (2)

Let $S=\mathrm{Spec} k[\varepsilon]/\varepsilon^2$ and $X=S\times T$ for some reduced $T$. Then $\mathscr O_X$ is not $S_1$ but $(\mathscr O_X)_{(\varepsilon)}\simeq \mathscr O_T$ is $S_1$.

This is not true.

(2) $\not\Rightarrow$ (1)

Let $f=\mathrm{id}_X$, then $F_s$ is supported at a point, so no matter what, $F_s$ is not $S_1$. So, you just have to choose an $X$ and and $F$ that's $S_1$.

(1) $\not\Rightarrow$ (2)

Let $S=\mathrm{Spec} k[\varepsilon]/\varepsilon^2$ and $X=S\times T$ for some reduced $T$. Then $\mathscr O_X$ is not $S_1$ but $(\mathscr O_X)_{(\varepsilon)}\simeq \mathscr O_T$ is $S_1$.

In general, you should expect that restrictions to fibers of morphisms are $S_n$ for a smaller $n$ than they are to start with. More precisely the following is true:

Let $f:X\to S$ be a flat morphism such that $S$ is an integral regular $1$-dimensional scheme and let $\mathscr F$ be a coherent sheaf on $X$. Then $\mathscr F$ is $S_n$ if and only if $\mathscr F_s$ is $S_{n-1}$ for every $s\in S$ closed point.

Remark: This is true for more general morphisms, but you can probably figure that out if you need to.

This is not true.

(2) $\not\Rightarrow$ (1)

Let $f=\mathrm{id}_X$, then $F_s$ is supported at a point, so no matter what, $F_s$ is not $S_1$. So, you just have to choose an $X$ and and $F$ that's $S_1$.

This is not true.

(2) $\not\Rightarrow$ (1)

Let $f=\mathrm{id}_X$, then $F_s$ is supported at a point, so no matter what, $F_s$ is not $S_1$. So, you just have to choose an $X$ and and $F$ that's $S_1$.

(1) $\not\Rightarrow$ (2)

Let $S=\mathrm{Spec} k[\varepsilon]/\varepsilon^2$ and $X=S\times T$ for some reduced $T$. Then $\mathscr O_X$ is not $S_1$ but $(\mathscr O_X)_{(\varepsilon)}\simeq \mathscr O_T$ is $S_1$.