This is my attempt: To show that $\sigma:E\rightarrow M$ where $\sigma$ is a $F$-algebra homomorphism is a linear transformation, we show that:
(i) $\forall a,b\in E, \sigma(a+b)=\sigma(a)+\sigma(b)$:
As $\sigma$ is a homomorphism, by the property of homomorphisms, $\forall a,b\in E, \sigma(a+b)=\sigma(a)+\sigma(b)$.
(ii)$\forall r\in F, a\in E, \sigma(r_Ea)=r_M\sigma(a)$
As $\sigma$ is a homomorphism, by the property of homomorphisms, $\sigma(r_{E}a)=\sigma(r_{E})\sigma(a)$.
As $\sigma$ is a $F$-algebra homomorphism, consider the homomorphisms $i_1:F\rightarrow E ,i_2:F\rightarrow M $we have $\sigma ∘i_1=i_2$. Let $r_E,r_M$ denote the image of $r\in F$ in $E$ and $M$ respectively, we have $\sigma(i_1(r))=\sigma(r_E)=i_2(r)=r_M$.
Thus we have $\forall r\in F, a\in E, \sigma(r_Ea)=r_M\sigma(a)$
I am not so sure of the definition of $F$ linear, and I am not able to find a clear and exact definition so far. Could someone please have a look to the proof to see if that is correct? Thanks in advance!
As someone said in the comments, a morphism of $F$-algebras is $F$ linear by definition. In fact, many places will define a morphism of $F$-algebras something like "A linear map $\sigma: E\to E $ is called a $homomorphism$ $of$ $F$-$algebras$ if ...". So, in other words, it is built in.
Generally speaking, in abstract algebra, a morphism preserves all of the relevant properties of the underlying algebraic structures. So, a ring homomorphism preserves all ring-like properties (addition and multiplication), and a group homomorphism preserves all group-like properties (the group operation), and similarly an $F$-algebra homomorphism preserves all $F$-algebra properties. Since an $F$-algebra is an $F$-vector space, a homomorphism preserves the linear structure, i.e., it is a linear map.