Revised Question:
From what I read, an algebra is locally finite if every finitely generated sub-algebra is finite, and a variety is locally finite if every member is locally finite (where a variety is the class of algebras closed under homomorphisms, sub-algebras, and direct products (direct product is denoted as $\prod_{i \in I}\mathbb{A}_i$)).
A variety is finitely generated if it is generated by a finite set of finite algebras.
Now, I read that there's a result (from universal algebra) that every finitely generated variety is locally finite.
The result says that given a variety generated by a set of finite algebras $\{\mathbb{A}_1,\mathbb{A}_2,...,\mathbb{A}_k\}$, we have that for any algebra $\mathbb{B}$ in the variety, there exists a homomorphism $\mathbb{C} \to \mathbb{B}$, for some algebra $\mathbb{C}$ in the variety, where $\mathbb{C} \cong \prod_{i\in I} \mathbb{C}_i$.
The proof uses the fact that $\textbf{Hom}(\mathbb{C},\mathbb{A}_i)$ is supposed to be a finite set - which is then supposed to imply that there are only finitely many maps from $\mathbb{C} \rightarrow \prod_{i \in I} \mathbb{C}_i\rightarrow \mathbb{C}_i$. Given this, there will be a finite subset $J$ of $I$ such that $\mathbb{C}$ embeds in $\prod_{j \in J}\mathbb{C}_j$, where $\prod_{j \in J}\mathbb{C}_j$ is a finite product of finite algebras. From this embedding of $\mathbb{C}$ into the finite product of finite algebras $\prod_{j \in J}\mathbb{C}_j$, we have that $\mathbb{C}$ is finite, and in turn, $\mathbb{B}$ is finite as well
But the proof did not go into any details as to why (1) $\textbf{Hom}(\mathbb{C},\mathbb{A}_i)$ is finite and how (2) this fact implies that $\mathbb{C} \to \prod_{i \in I}\mathbb{C}_i \to \mathbb{C}_i$ ... Any help for (1) and (2) ?
First, let me include the full proof from the book you're most likely using. It'll be easier than rewriting it.
The book is Bergman's "Universal Algebra".
First, you stated that $\mathbb{C}$ is isomorphic to $\prod_{i\in I}\mathbb{C}_i$. This is not true, it's only isomorphic to a sub-algebra of $\prod_{i\in I}\mathbb{C}_i$.
Since $X$ generates $\mathbb{C}$, $C = \bigcup_{n=1}^\infty X_n$ where $X_0 = X$, $X_{n+1} = X_n\cup \{ p(a_1, ..., a_k): a_i\in X_n, p\text{ is } k\text{-ary}\}$.
Now if two homomorphisms $f, g:\mathbb{C}\to\mathbb{A}_i$ are equal on $X$, one can prove (and this is easy) by induction that they are equal on each of $X_n$, so that they are equal on all of $C$, so $f=g$.
Therefore the map $f\mapsto f|_X, \text{Hom}(\mathbb{C}, \mathbb{A_i})\to A_i^X$ which sends a homomorphism to its restriction to $X$ is injective. So there can be at most $|A_i^X|<\infty$ of such homomorphisms.
I hope this clears it up, I invite you to a further discussion if it's not clear.