Denote the basis of $E_0$ (resp. $E_1$, $S_0$, $S_1$) (sorry that I changed the notation because I do not want to have too many f's) by $e_{0,1},\cdots,e_{0,n_0}$ (resp. $e_{1,1},\cdots,e_{1,n_1}$, $s_{0,1},\cdots,s_{0,m_0}$, $s_{1,1},\cdots,s_{1,m_1}$, and the matrix corresponds to $f_0$ (resp. $f_1$) by $F_0$ (resp. $F_1$). It is always possible to choose bases of $E_1$ and $S_1$ such that the matrices corresponding to differentials are of the form
$$\begin{pmatrix}
I & 0
\end{pmatrix}$$
or its transpose.
Case 1: if $f$ is split injective, then there is an exact sequence
$$0\rightarrow E_.\rightarrow S_. \rightarrow H_. \rightarrow 0$$
with $H_.$ acyclic. Then the determinant of $f$ can be obtained by the combine the isomorphisms $\det E_. \otimes\det H_. \xrightarrow{\sim} \det S_.$ and $\det S_. \xrightarrow{\sim} \mathbb{1}$.
In this case, you just have invertible matrices $F_0$ and $F_1$, and the basis of $S_.$ can be chosen suitable such that the first half is mapped from $E_.$ and the lower half of basis of $S_.$ contributes to the basis of $H_.$. Also, the map from $S_.$ to $H_.$ is projection. By above, $\det f$ is just a scaler multiple whose scaler is given by $\det F_0 (\det F_1)^{-1}$.
Case 2: otherwise, we can construct an intermediate complex, namely $C_.$, which is defined as $$C_i:=E_i\oplus S_i \oplus E_{i-1}$$ with differentials
$$d_{C,i}:=
\left( \begin{array}{ccc}
d_{E,i} & 0 & -1\\
0 & d_{S,i} & f_{i+1} \\
0& 0& -d_{E,i-1}
\end{array} \right)$$
and the maps
$$
\begin{aligned}
&\alpha: E_. \rightarrow C_., e \mapsto (e,0,0)\\
&\beta: S_.\rightarrow C_., s \mapsto (0,s,0)\\
&\gamma: C_.\rightarrow S_., (e,s,e') \mapsto f(e)+s.
\end{aligned}
$$
Then it is easy to check that $\gamma\beta=\operatorname{id}.$ and $\gamma\alpha=f$. So we can apply the method in case 1 to construct $\det f$, which is $\det^{-1}\beta \det\alpha$.
In your case, by choosing the bases wisely, the only non-trivial part during computing $\det f$ is the determinant of $\alpha$, which is also not too complicated when written as matrix. And the final conclusion is that $$\det f=\det F_0(\det F_1)^{-1}.$$
$\DeclareMathOperator\Hom{Hom}$The answer seems to be positive.
Let $\psi_i: X_i \rightarrow Y_i$. Consider the directed system of exact sequences
$$ 0 \longrightarrow K_i \longrightarrow X_i\otimes- \xrightarrow{\psi_i \otimes -} Y_i \otimes -$$
in the functor category. Its direct limit equals an exact sequence
$$ 0 \longrightarrow K \longrightarrow X\otimes- \xrightarrow{\psi \otimes -} Y \otimes -$$
since direct limits are exact and tensors commute with them. Now $M$ is pure-injective, so the functor $M\otimes-$ is injective and there exists an inverse system of short exact sequences
$$ \Hom(Y_i\otimes-,M\otimes-) \xrightarrow{\Hom(\psi _i \otimes - , M\otimes -)} \Hom(X_i\otimes-, M\otimes-)\longrightarrow \Hom(K_i, M \otimes-)\longrightarrow 0. $$
The natural isomorphism $\Hom(\psi _i \otimes - , M\otimes -)\cong \Hom(\psi_i, M)$ implies $\Hom(K_i, M \otimes-) = 0$. Because $\Hom(-,-)$ turns direct limits in the first argument into inverse limits, taking the inverse limit of the inverse system yields
$$ \Hom(Y\otimes-,M\otimes-) \xrightarrow{\Hom(\psi \otimes - , M\otimes -)} \Hom(X\otimes-, M\otimes-)\longrightarrow \Hom(K, M \otimes-) = 0.$$
Hence $\Hom(\psi, M)\cong \Hom(\psi \otimes - , M\otimes -) $ is surjective.
Best Answer
This isn’t true when $R=\mathbb{Z}$, $F=\mathbb{Q}$, $M=\mathbb{Z}$ and $I^\bullet$ is the complex $$\cdots\to0\to\mathbb{Z}\to\mathbb{Q}\to\mathbb{Q}/\mathbb{Z}\to0\to\cdots$$ with $\mathbb{Z}$ in degree zero.
Then $\varphi$ is a map from $\mathbb{Q}$, as a complex concentrated in degree zero, to the zero complex.