$\DeclareMathOperator{\im}{Im}$
Let's do some grunt diagram chasing.
First, we prove $\overline{\delta}$ is injective. Suppose $x\in\ker\overline{\delta}$. Then $x\in\ker(\delta)=\im(\varepsilon)$, so $x=\varepsilon(k)$ for some $k\in K$. But also, $x\in\ker\varphi$, so
$$0=\varphi(x)=\varphi(\varepsilon(k))=\varepsilon'(\alpha(k))$$
But $\epsilon'$ is injective, so $\alpha(k)=0$. Since $\alpha$ is an isomorphism $k=0$, so $x=\varepsilon(k)=\varepsilon(0)=0$.
Therefore, $\overline{\delta}$ is injective.
Now let us prove $\overline{\delta}$ is surjective. Let $y\in\ker\psi$. Then $\psi(y)=0$, so
$$0=\eta'(\psi(y))=\beta(\eta(y))$$
Since $\beta$ is an isomorphism, $\eta(y)=0$, so $y\in\ker(\eta)=\im(\delta)$. Write $y=\delta(x)$ for some $x\in M$.
We have $\delta'(\varphi(x))=\psi(\delta(x))=\psi(y)=0$, so $\varphi(x)\in\ker\delta'=\im(\varepsilon')$. Let $k'\in K'$ such that $\varphi(x)=\varepsilon'(k')$. Since $\alpha$ is an isomorphism, there is $k\in K$ such that $k'=\alpha(k)$. Then
$$\varphi(\varepsilon(k))=\varepsilon'(\alpha(k))=\varepsilon'(k')=\varphi(x)$$
So now we use the $R$-module structure: Let $z=x-\varepsilon(k)$. Then the above means that $z\in\ker\varphi$. We prove that $\overline{\delta}(z)=y$:
$$\overline{\delta}(z)=\delta(x)-\delta\varepsilon(k)=\delta(x)=y$$
because $\delta\varepsilon=0$, since the first row is exact.
Equivalently, you are asking about the restriction map $\operatorname{Hom}(P, M) \to \operatorname{Hom}(Q, M)$ induced by the inclusion map $Q\to P$ where $P=\prod_s N_s$ and $Q=\bigoplus_s N_s$. This map is part of an exact sequence $$0\to \operatorname{Hom}(P/Q,M)\to\operatorname{Hom}(P,M)\stackrel{\phi}\to\operatorname{Hom}(Q,M)\to \operatorname{Ext}^1(P/Q,M)\to \operatorname{Ext}^1(P,M).$$ So, $\phi$ is injective iff there are no nontrivial homomorphisms $P/Q\to M$. In particular, for instance, this means that $\phi$ is never injective when $R$ is a field as long as infinitely many of the $N_s$ are nontrivial and $M$ is nontrivial. Or, for an arbitrary family where infinitely many of the $N_s$ are nontrivial, if $M=P/Q$ then $\phi$ will not be injective. On the other hand, if $M$ is an injective module (so in particular, if $R$ is a field, or more generally if $R$ is a semisimple ring), then $\phi$ is always surjective.
For an example where $\phi$ is not surjective, you could take $R=\mathbb{Z}$ and consider the direct product $P=\prod_p\mathbb{Z}/(p)$ where $p$ ranges over the primes. Note that then the quotient $P/Q$ is divisible: given an element $x\in P$ and a nonzero integer $n$, you can divide $x$ by $n$ mod $Q$ by just ignoring the coordinates corresponding to the finitely many primes dividing $n$. But $P$ has no nontrivial divisible submodules, since if $x\in P$ is nonzero on its $p$th coordinate then $x$ is not divisible by $p$. So in particular, $P/Q$ is not isomorphic to a submodule of $P$ so the inclusion map $Q\to P$ does not split. Taking $M=Q$, this says exactly that $\phi$ fails to be surjective since its image does not contain the identity map $Q\to Q$.
To get an example where $\phi$ is neither surjective nor injective, you can just take a direct sum of examples of each type. For instance, with the same product as in the previous example, you could take $M=Q\oplus P/Q$.
Another well-known example where $\phi$ is not surjective is if you take $R=\mathbb{Z}$ and $P$ to be a product of countably infinitely many copies of $\mathbb{Z}$. Then it is a nontrivial theorem that every homomorphism $P\to\mathbb{Z}$ is a linear combination of the projection maps. In particular, there are only countably many such maps, whereas there are uncountably many homomorphisms $Q\to\mathbb{Z}$ (since $Q$ is free on infinitely many generators), so the restriction map $\phi:\operatorname{Hom}(P,\mathbb{Z})\to\operatorname{Hom}(Q,\mathbb{Z})$ is very far from surjective. (On the other hand, it follows from the theorem that $\phi$ is injective in this case, which is rather surprising.)
Best Answer
First of all, for $\varphi$ to be a homomorphism, it means that $\varphi(fr + gs) = \varphi(f)r + \varphi(g)s$ for $r, s \in R$ and $f, g \in \operatorname{Hom}_R(R,M)$. Notice that multiplication is not an operation on modules, only addition and scalar multiplication.
Secondly, it is not clear to me why $f(1) = g(1) \implies f = g$. But we can handle this and surjectivity at the same time with the following realization.
To see this, define $f(r) = mr$. Then clearly $f(1) = m$. If there is another $R$-linear map $g$ with $g(1) = m$, then $mr = g(1)r = g(r)$ for all $r \in R$ so $g = f$ .