EDIT: My previous version of this was not rigorous enough. I was trying to be clever and get away with just simple cell attachments, which only work if you already know that the functor is represented by a space. Sorry for the delay in reworking, but this particular proof has enough details that it takes time to write up.

As you say, you begin by decomposing such functors so without loss of generality F(pt) $F(pt)$ is a single point.

Start with X-1 = pt$X_{-1}$ as a point. Assume you've inductively constructed an n-dimensional $(n-1)$-dimensional complex Xn-1$X_{n-1}$ with an element xn-1 F(Xn-1) $x_{n-1} \in F(X_{n-1})$ so that, for all k < n, [Sk,Xn-1] F(Sk) is a bijection.

Take YnCW-inclusions $Z \to be the wedge Y$ of Xn-1finite CW complexes with $Y$ formed by attaching a collection of copies of Sn$k$-cell for $k < n$, one per element of F(Sn) not in the image of map$$[Sn,Xn-1]. The Mayer-Vietoris property implies there exists an element y F(Yn) restricting Y,X_{n-1}] \to xn-1 on X and so that the map [Sn,Yn] F(SnZ,X_{n-1}] \times_{F(Z)} F(Y)$$is surjective.Also, [Sk,Yn] [Sk,Xn-1] is an isomorphism for k < n.

Now, define a "problem" of dimension $n$ to be a CW-inclusion $Z \to Y$ where $Y$ is a subspace of $\mathbb{R}^\infty$ formed by attaching a single $n$-cell to $Z$, together with an element of[Sn v Sn,Yn] whose image in F(Sn v Sn$$Map(Z,X_{n-1}) is \times_{F(Z)} F(Y).$$The fact that $Y$ has a fixed embedding in $\mathbb{R}^\infty$ means that there is a set of problems $S$, whose elements are tuples $(Z_s,Y_s,f_s,y_s)$ with $f_s$ a map $Z_s \to X_{n-1}$ and $y_s$ is a compatible element of $F(Y)$.

Let $X_n$ be the image pushout of F(Sn) under the fold map. If C = diagram$$X_{n-1} \leftarrow \coprod_{s \in Sn x [0,1] / } Z_s \rightarrow \coprod_{s \in Sn x {0}, then } Y_s$$where the Mayer-Vietories property implies that lefthand maps are defined by the "problems" maps $f_s$ and the righthand maps are in bijection with the elements of F(Sn v Sn) that lift to F(C). Note that C given $CW$-inclusions. This is the union of Sn v Sn with a single (n+1)-cell.

Let Xn be relative $CW$-inclusion formed by attaching a collection of (n+1)-cells, one per problem$n$-cells; therefore, using pushout diagrams

C Sn v Sn Yn

defined by $X_n$ still has the problems. The Mayer-Vietoris extension property implies that yn lifts for relative cell inclusions of dimension less than $n$.

The space $X_n$ is homotopy equivalent to an element xn F(Xn)the homotopy pushout of the given diagram, which is formed by gluing together mapping cylinders. I claim that [SnSpecifically, Xn] F(Sn) $X_n$ is an isomorphism; weakly equivalent to the proof is that [Sn,Yn] space$$X_{n-1} \times \{0\} \cup (\coprod_S Z_s \times [Sn,Xn] is a surjection because we've only attached 0,1]) \cup (n+1)-cells, and if you have \coprod_S Y_s \times \{1\})$$which decomposes into two elements in CW-subcomplexes:$$A = X_{n-1} \times \{0\} \cup (\coprod Z \times [Sn0,1/2])$$which deformation retracts to $X_{n-1}$, Yn] having and$$B = (\coprod Z_s \times [0,1/2]) \cup (\coprod Y_s \times \{1\})$$which deformation retracts to $\coprod Y_s$ with intersection $A \cap B \cong \coprod Z_s$. The Mayer-Vietoris property and the same image in F(Sn) coproduct axiom then you can use the Mayer-Vietories property to explicitly construct a "problem" which has been fixed imply that there is an element $x_n \in XnF(X_n)$ whose restriction to $A$ is $x_{n-1}$ and whose restriction to $B$ is $\prod y_s$.

Taking colimits, you have a space X CW-complex $X$ with an element $x F(X) \in F(X)$ (constructed using a mapping telescope + Mayer-Vietoris argument) so that, for all kCW-inclusions $Z \to Y$ obtained by attaching a single cell, the map$$[Sk,X] F(SnY,X] \to [Z,X] \times_{F(Z)} F(Y)$$ is a bijectionsurjective.

Now you need to show that for any finite CW-complex K, [K,X] F(X) $K$, $[K,X] \to F(X)$ is a bijection. Without loss of generality we can assume K is connected and has only a single zero-cell.

First, surjectivity is straightforward by induction on the skeleta of K. $K$. More specifically, for any K $K$ with subcomplex L, $L$, element of F(K), $F(K)$, and map $L X \to X$ realizing the restriction to F(L), $F(L)$, you induct on the cells K\L. Each cell has an attaching map coming from the trivial element of F(Sk), and so you can extend the map to this new cell$K\setminus L$. (The fact that you're using free homotopy classes of maps forces you to be slightly more cautious, but free homotopy classes of maps from a sphere are just orbits of based maps under the action of π1 and so you can still safely talk about "trivial" element.)Then, injectivity: if you have two elements $K X \to X$ with the same images in F(K), $F(K)$, you use the above-proven stronger surjectivity property to the inclusion $K x {0,1} \times \{0,1\} \to K x \times [0,1] 0,1]$ to show that there is a homotopy between said maps.

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Yes, Brown representability holds for such functors. There are not really any material differences between this and the proof of Brown representability in the pointed case.

As you say, you begin by decomposing such functors so without loss of generality F(pt) is a single point.

Start with X-1 = pt. Assume you've inductively constructed an n-dimensional complex Xn-1 with an element xn-1 ∈ F(Xn-1) so that, for all k < n, [Sk,Xn-1] → F(Sk) is a bijection.

Take Yn to be the wedge of Xn-1 with a collection of copies of Sn, one per element of F(Sn) not in the image of [Sn,Xn-1]. The Mayer-Vietoris property implies there exists an element y ∈ F(Yn) restricting to xn-1 on X and so that the map [Sn,Yn] → F(Sn) is surjective. Also, [Sk,Yn] → [Sk,Xn-1] is an isomorphism for k < n.

Now, define a "problem" to be an element of [Sn v Sn,Yn] whose image in F(Sn v Sn) is in the image of F(Sn) under the fold map. If C = Sn x [0,1] / Sn x {0}, then the Mayer-Vietories property implies that the "problems" are in bijection with the elements of F(Sn v Sn) that lift to F(C). Note that C is the union of Sn v Sn with a single (n+1)-cell.

Let Xn be formed by attaching a collection of (n+1)-cells, one per problem, using pushout diagrams

C ← Sn v Sn → Yn

defined by the problems. The Mayer-Vietoris property implies that yn lifts to an element xn ∈ F(Xn). I claim that [Sn,Xn] → F(Sn) is an isomorphism; the proof is that [Sn,Yn] → [Sn,Xn] is a surjection because we've only attached (n+1)-cells, and if you have two elements in [Sn,Yn] having the same image in F(Sn) then you can use the Mayer-Vietories property to explicitly construct a "problem" which has been fixed in Xn.

Taking colimits, you have a space X with an element x ∈ F(X) (constructed using a mapping telescope + Mayer-Vietoris argument) so that, for all k, [Sk,X] → F(Sn) is a bijection. Now you need to show that for any finite CW-complex K, [K,X] → F(X) is a bijection. Without loss of generality we can assume K is connected and has only a single zero-cell.

First, surjectivity is straightforward by induction on the skeleta of K. More specifically, for any K with subcomplex L, element of F(K), and map L → X realizing the restriction to F(L), you induct on the cells K\L. Each cell has an attaching map coming from the trivial element of F(Sk), and so you can extend the map to this new cell. (The fact that you're using free homotopy classes of maps forces you to be slightly more cautious, but free homotopy classes of maps from a sphere are just orbits of based maps under the action of π1 and so you can still safely talk about "trivial" element.)

Then, injectivity: if you have two elements K → X with the same images in F(K), you use the above-proven stronger surjectivity property to the inclusion K x {0,1} → K x [0,1] to show that there is a homotopy between said maps.