Let $M_i$ submodule of a module $M$ (over $R$). Do we have $$\bigoplus_{i=1}^d M_i\cong \sum_{i=1}^d M_i\ \ ?$$
I think yes, and my proof goes as follow.
Let \begin{align*} \Phi: \sum_{i=1}^d M_i&\longmapsto \bigoplus_{i=1}^dM_i\\ m_1+...+m_d&\longmapsto (m_1,...,m_d). \end{align*} This is clearly a $R-$module morphism. The surjection is obvious. For the injection, $$0=\Phi(m_1+...+m_d)=(m_1,...,m_d)=0\implies m_i=0,$$ and thus $\Phi$ is bijective what prove the claim.
Is it correct ? Does it also work if $d=+\infty $ ? If yes, $$\sum_{i=1}^\infty M_i=\{m_1+m_2+...\mid\text{ where $m_i\neq 0$ for only a finite number of }m_i\}\ ?$$
Because in my course, the definition of semi-simple module $M$ is if $$M=\sum_{i=1}^d M_i$$ where $M_i$ are simple. Then I have a remark that say that we can take the sum infinite, and then, I have a proposition that say that $M$ is simple iff $M=\bigoplus_{i=1}^d M_i$ where $M_i$ are simple. That's why I'm confused.
No, the proof is incorrect.
There is a surjective homomorphism the other way around: $$ \sigma\colon\bigoplus_{i=1}^d M_i\to\sum_{i=1}^d M_i, \qquad \sigma(x_1,x_2,\dots,x_d)=x_1+x_2+\dots+x_d, $$ which in general is not an isomorphism.
For instance, with $d=2$ and $M_1=2\mathbb{Z}$, $M_3=3\mathbb{Z}$ as submodules of $\mathbb{Z}$. Then the direct sum is isomorphic to $\mathbb{Z}^2$, whereas the sum is isomorphic to $\mathbb{Z}$.
Your attempt at defining a map is incorrect, because an element $x\in\sum_i M_i$ can be represented in distinct ways as $$ x=x_1+x_2+\dots+x_d=x_1'+x_2'+\dots+x_d' $$ with $x_i,x_i'\in M_i$, for $i=1,2,\dots,d$.
Just think to vector spaces and you'll immediately find other examples.
As another example, $\mathbb{R}^2$ has infinitely many subspaces of dimension $1$. Their sum is just $\mathbb{R}^2$ and is surely not direct.