$
\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:
- For any associative algebra $A$ and linear $f : X \to A$ such that $f(x)^2 = Q(x)$ for all $x \in X$, there is a unique algebra homomorphism $F : \Cl(X, Q) \to A$ such that $F(x) = f(x)$ for all $x \in X$.
Choose $Q = \hat fq$ and $A = \Cl(Y, q)$ and simply widen the codomain of $f : X \to Y$ to get $f : X \to \Cl(Y, q)$.
