First of all, $\text{Aff}\left ( \mathbb{R}^n ,\mathbb{R}^n \right )$ is a set of affine maps $\mathbb{R}^n \to \mathbb{R}^n$. And, $M(n,\mathbb{R})$ is a set of all $n \times n$ matrices over $\mathbb{R}$.
Let $L_f$ be a linear map that corresponds to $f \in \text{Aff}\left ( \mathbb{R}^n ,\mathbb{R}^n \right )$, which is defined by $L_f(x)=f(x)-f(0)$.
I have to show that $L: \text{Aff}\left ( \mathbb{R}^n ,\mathbb{R}^n \right ) \to M(n,\mathbb{R})$ is a map of rings.
So far, I've checked $\text{Aff}\left ( \mathbb{R}^n ,\mathbb{R}^n \right )$ is a ring under addition and composition.
Also, it was easy to check that $L_{f_1+f_2}(x)=(L_{f_1}+L_{f_2})(x)$, and the identity preserving axiom.
However, I'm stuck showing for the other axiom.
i.e.: $L_{f_1 \circ f_2}(x)=(f_1(x)-f_1(0)) \circ (f_2(x)-f_2(0))=^?(f_1 \circ f_2)(x)-(f_1 \circ f_2 )(0)$
How to show this? Should I just change all $\circ$'s on the RHS to matrix multiplication?
Any help would be appreciated a lot!
Actually, your $L$ sends an affine map to a linear map, to get a matrix you need to choose a basis, in this situation there is the standard one in $\mathbb{R}^n$.
So $L_{f_1 \circ f_2}(x)=(f_1 \circ f_2)(x)-(f_1 \circ f_2 )(0)=$ $=f_1( f_2(x))-f_1(f_2 (0)).$
By definition of affine map, this is exactly the linear map associated to $f_1$ evaluated on the vector $ f_2(x)-f_2(0)$, which is exactly $(f_1(x)-f_1(0)) \circ (f_2(x)-f_2(0))$.
Finally, the matrix associated to the composition of linear maps is the product matrix so you get the claim.
If you prefer, you also choose an origin, again there is a standard one, then $f_i(x)=A_ix+b_i$ for some matrix $A_i$ and some vector $b_i$ and the result follows by explicit computations.