In Hoffman and Kunz section 3.2, they prove that $L(V,W)$ is a vector space where $V,W$ are vector spaces over the same field $F$(theorem 4). Then in theorem 5, they prove that this space is finite dimensional and its dimension is $dim(V)\times dim(W)$. Then the text gives a theorem with a lemma showing that $L(V)$ has the following properties for every $U,T,T_1,T_2\in L(V,V)$, for every $c\in F$.
(i) $IU=UI=U$
(ii) $U(T_1+T_2)=UT_1+UT_2$; $(T_1+T_2)U=T_1U+T_2U$
(iii) $c(UT)=(cU)T=U(cT)$
Then the authors remark that this lemma together with a portion of theorem 5 tell us that $L(V,V)$ together with composition is a linear algebra with identity.
But before theorem 5, we have established that $L(V,V)$ is a vector space over $F$ and in the lemma, we show that it satisfies the additional conditions required from a linear algebra with identity. So i have no idea why we need a portion of theorem 5 here.
So my question is why do we need theorem 5 to prove that $L(V,V)$ is a linear algebra with identitiy?
You are right: it is not needed. My guess is that they meant theorem 6, which proves (among other things) that if you compose two linear functions, you get again a linear function.