Assume that $(X, \mathbb{R}), (Y, \mathbb{R})$ are a vector space over $\mathbb{R}$, together with a linear map $f : X\rightarrow Y$ and pullback quadratic form $(\hat{f}q)(x) = q(f(x))$ for every $x\in X$.
Prop: There exists an algebra homomorphism $F : Cl(X, \hat{f}q)\rightarrow Cl(Y, q)$.
If $\{e_i\}_{1\leq i\leq n}\subseteq X$ is an orthonormal basis in $X$, then by using the fact that for every $i\neq j$,
$$\{e_{i}e_{j}\} = e_{i}e_{j} + e_{j}e_{i} = 0$$
we conclude that
$$\mathfrak{cl}(X, q) = \{e_{i_1}e_{i_2}\cdots e_{i_k} : i_1< \cdots < i_{k}\wedge k\in \{1, ..., n\}\}\subset Cl(X, \hat{f}q)$$
is a basis in $Cl(X, \hat{f}q)$. Then for each $n$-product $e_{i_1}e_{i_2}\cdots e_{i_n}\in Cl(X, \hat{f}q)$, we can define a linear map
$$f_n : e_{i_1}e_{i_2}\cdots e_{i_n}\in \text{cl}(X, \hat{f}q)\mapsto f(e_{i_1})f(e_{i_2})\cdots f(e_{i_n})\in \text{cl}(Y, q)$$
However, I am not really sure if I am on the right track. Could I hear your feedback, if possible?
$ \newcommand\Cl{\mathrm{Cl}} \newcommand\m\mathbf $If you remove the denominators from your $F$ then you're on the right track, you just need to show that $f$ as defined is an algebra homomorphism. I think that you're overcomplicating things by introducting $\Cl^j$.
I will just use juxstaposition for the algebra products. You're expression for a basis of $\Cl(X, \hat fq)$ is not quite right; it should be $$ B = \{e_{i_1}\dotsb e_{i_k} \;:\; i_1 < \dotsb < i_k,\: k = 1,\dotsc,n\} $$ where $n$ is the dimension of $X$. Notice the $<$ rather than $\leq$. It also makes no sense to say $$ e_i^2 = -1 $$ for several reasons. First, if e.g. $\mathbb F = \mathbb R$ then the sign of $e_i^2$ depends on the quadratic form $\hat fq$. Second, you cannot normalize basis over an arbitary field $\mathbb F$; it just isn't something that makes sense. (Hint: consider the square classes $\mathbb F^\times/\mathbb F^{\times2}$.) What we can do is always find an orthogonal basis so that $$ e_ie_j + e_je_i = 0,\quad i \ne j $$ as you say, and otherwise we will just say $$ e_i^2 = (\hat fq)(e_i) = q(f(e_i)) $$ which is by definition of $\Cl(X, \hat fq)$.
What follows will be much easier using multi-index notation. A multi-index $\m i$ is a tuple $\m i = (i_1,\dotsc, i_m)$ such that $1 \leq i_1 < \dotsb < i_m$. We define $$ |\m i| = m,\quad \m i \leq p \iff i_m \leq p,\quad p \in \m i \iff \exists q.\;p = i_q, $$$$ e_{\m i} = e_{i_1}\dotsb e_{i_m},\quad g(e)_{\m i} = g(e_1)\dotsb f(e_{i_m}). $$ Here $g$ is any function. Using the above definition of $\in$ we apply set-like operations $\cap, \cup, \setminus$ to multi-indices by adding the stipulation that the result be a well-defined multi-index.
In multi-index notation we may write $$ B = \{e_{\m i} \;:\; \m i \leq n\}. $$ Now we can define a linear map $F : \Cl(X, \hat fq) \to \Cl(Y, q)$ via $$ F(e_{\m i}) = f(e)_{\m i} $$ All we need to do now is show that $F$ is a homomorphism. By linearity it suffices to show that $$ F(e_{\m i})F(e_{\m j}) = F(e_{\m i}e_{\m j}). \tag{$*$} $$ Note that by definition of $\Cl(X, \hat fq)$ and $\Cl(Y, q)$ $$ f(v)^2 = q(f(v)) = (\hat fq)(v) = v^2. $$ In particular $f(e_l)^2 = e_l^2$ for any $l$. It also follows that $f(e_l)$ and $f(e_m)$ anticommute when $l \ne m$: $$\begin{aligned} 0 &= e_le_m + e_me_l \\ &= (e_l + e_m)^2 - e_l^2 - e_m^2 \\ &= f(e_1+e_m)^2 - f(e_1)^2 - f(e_m)^2 \\ &= f(e_l)f(e_m) + f(e_m)f(e_l). \end{aligned}$$ Thus we now calculate $$\begin{aligned} F(e_{\m i})F(e_{\m j}) &= f(e)_{\m i}f(e)_{\m j} \overset1= \pm[f(e)^2]_{\m i\cap\m j}f(e)_{\m i\cup\m j} \\ &= \pm(e^2)_{\m i\cap\m j}f(e)_{\m i\cup\m j} = \pm F\Bigl((e^2)_{\m i\cap\m j}\,e_{\m i\cup\m j}\Bigr) \\ &\overset2= F(e_{\m i}e_{\m j}) \end{aligned}$$ which is ($*$). I have opted to not write out the sign $\pm$ we get in (1) from (anti)commuting basis vectors; the exact inverse set of (anti)commutations is performed in (2), so the sign cancels.
Alternatively, this is trivial if you've already proved the universal propery of Clifford algebras:
Choose $Q = \hat fq$ and $A = \Cl(Y, q)$ and simply widen the codomain of $f : X \to Y$ to get $f : X \to \Cl(Y, q)$.