$\newcommand{\Hom}{\mathrm{Hom}}$
Let $K,M$ and $N$ be $R$-modules. Show that $$\Hom_R(K,M\oplus N)\cong \Hom_R(K,M)\oplus \Hom_R(K,N).$$
Let $\pi_M:M\oplus N\longrightarrow M$ and $\pi_N:M\oplus N\longrightarrow N$ be the projection, i.e. $\pi_M(m,n)=m$ and $\pi_N(m,n)=n$. Let \begin{align*} \Phi:\Hom_R(K,M\oplus N)&\longrightarrow \Hom_R(K,M)\oplus \Hom_R(K,N)\\ \varphi&\longmapsto (\pi_M\circ \varphi,\pi_N\circ\varphi). \end{align*}
I'm trying to show that $\Phi$ is an isomorphism. $$\Phi(\varphi)=0\implies \forall k\in K, \Phi(\varphi)(k)=0\implies \forall k\in K,(\pi_M(\varphi(k)),\pi_N(\varphi(k)))=(0,0).$$ Let $\varphi(k)=(\varphi_1(k),\varphi_2(k))\in M\oplus N$. Then, we have that $$(\pi_M(\varphi(k)),\pi_N(\varphi(k))=(0,0)\implies \varphi_1(k)=\varphi_2(k)=0$$ and thus $\varphi_1=\varphi_2=0$ i.e. $\varphi=0$. Therefore $\Phi$ is one-to-one. Let $(\varphi,\mu)\in \Hom_R(K,M)\oplus \Hom_R(K,N)$. Then, if we set $\tau=(\mu,\varphi)$ then $\tau\in \Hom_R(K,M\oplus N)$ and $\Phi(\tau)=(\varphi,\mu),$ what conclude for the surjectivity.
Is it correct ? (I have doubt for the surjectivity).
— Your proof for the injectivity is fine. For surjectivity, I believe that your notation $\tau=(\mu,\phi)$ means that we define $\tau \in \mathrm{Hom}_R(K,M\oplus N)$ by \begin{align*} \tau: K &\longrightarrow M \oplus N\\ x&\longmapsto (\mu(x),\phi(x)). \end{align*} which is indeed correct. You just have to check that $\tau$ is indeed an $R$-modules morphism.
— However, you forgot to check that $\Phi$ is an $R$-modules morphism, that is: $$\forall r \in R, \quad \Phi(r \cdot \phi) = r \cdot' \Phi(\phi)$$ I let you think about it, this is not difficult.
– Notice that more generally we have $$\mathrm{Hom}_R\left(A ; \prod_{i \in I} B_i\right) \cong \prod_{i \in I} \mathrm{Hom}_R(A ; B_i)$$ as $R$-modules, where $R$ is a commutative ring (see Rotman, An introduction to homological algebra, theorem 2.30).