The problem seems to be that you're computing expectations without conditioning on the available evidence. More formally, some of the equations with implicit sums are incomplete. For example, here:
$$P(X_2|X_1)=E(P(X_2|X_1,S_1,S_2))?$$
That isn't quite right if taken literally:
$$P(X_2|X_1)=\sum_{s_1,s_2}P(X_2|X_1,S_1,S_2)P(S_1,S_2)?$$
It should be:
$$P(X_2|X_1)=\sum_{s_1,s_2}P(X_2|X_1,S_1,S_2)P(S_1,S_2|X_1).$$
Later, in the computations for general $n$, the unconditional probabilities $\boldsymbol{\pi}'=(P(S_{n-1}=1)P(S_{n-1}=2))$ shouldn't appear so early. Instead, we should see terms like $P(S_{n-1}=1|X_{n-1},\dots,X_1)$. (But I haven't worked out the full expression involving the latter.)
Edit: After discussion in the comments, I understand what you're attempting a little more clearly. The above issue was a bit of a red herring. There is an improper marginalization in the equation that defines your overall strategy:
$$P(X_2,\dots,X_n|X_1)=P(X_n|X_{n-1},\dots,X_1)\cdot P(X_{n-1}|X_{n-2},\dots,X_1)\cdot \dots \cdot P(X_2|X_1)?$$
For simplicity, take $n=3$:
$$P(X_2,X_3|X_1)=P(X_3|X_2,X_1) \cdot P(X_2|X_1)?$$
Now remember that each $P$ above is really a marginal over the $S_i$:
$$\sum_{S_1,S_2,S_3}P(X_2,X_3,S_1,S_2,S_3|X_1)=\sum_{S_1,S_2,S_3}P(X_3, S_1,S_2,S_3|X_2,X_1) \cdot \sum_{S_1',S_2',S_3'}P(X_2,S_1',S_2',S_3'|X_1)?$$
But that's not right. Applying the definition of $P(X_2,X_3|X_1)$ gives us only a single summation on the right side:
$$\sum_{S_1,S_2,S_3}P(X_2,X_3,S_1,S_2,S_3|X_1)=\sum_{S_1,S_2,S_3}P(X_3, S_1,S_2,S_3|X_2,X_1) \cdot P(X_2,S_1,S_2,S_3|X_1).$$
More to the point, we can apply similar reasoning to $P(X_1,X_2,X_3)$, eventually getting:
$$P(X_1,X_2,X_3)=\sum_{S_1,S_2,S_3}P(X_3|S_3)P(S_3|S_2)P(X_2|S_2)P(S_2|S_1)P(X_1|S_1)P(S_1).$$
CAVEAT: In the above equation and what follows, I assume that the $X_t$ are conditionally independent given $S_t$. This assumption makes the formulas more compact, but it isn't essential. To revoke the assumption, just replace $P(X_2|S_2)$ with $P(X_2|X_1,S_2)$ and so on.
And that's the matrix product given in the article! To build up the product manually, work from right to left:
$$P(S_1)
=\left(\begin{array}{c}P(S_1=1)\\P(S_1=2)\end{array}\right)
=\boldsymbol{\pi}
=\boldsymbol{P}'\boldsymbol{\pi}$$
$$P(X_1|S_1)P(S_1)=\boldsymbol{F}_1\boldsymbol{P}'\boldsymbol{\pi}$$
$$\sum_{S_1}P(S_2|S_1)P(X_1|S_1)P(S_1)=\boldsymbol{P}'\boldsymbol{F}_1\boldsymbol{P}'\boldsymbol{\pi}$$
$$\sum_{S_1}P(X_2|S_2)P(S_2|S_1)P(X_1|S_1)P(S_1)=\boldsymbol{F}_2\boldsymbol{P}'\boldsymbol{F}_1\boldsymbol{P}'\boldsymbol{\pi}$$
$$\sum_{S_1,S_2}P(S_3|S_2)P(X_2|S_2)P(S_2|S_1)P(X_1|S_1)P(S_1)=\boldsymbol{P}'\boldsymbol{F}_2\boldsymbol{P}'\boldsymbol{F}_1\boldsymbol{P}'\boldsymbol{\pi}$$
$$\sum_{S_1,S_2}P(X_3|S_3)P(S_3|S_2)P(X_2|S_2)P(S_2|S_1)P(X_1|S_1)P(S_1)=\boldsymbol{F}_3\boldsymbol{P}'\boldsymbol{F}_2\boldsymbol{P}'\boldsymbol{F}_1\boldsymbol{P}'\boldsymbol{\pi}$$
$$\sum_{S_1,S_2,S_3}P(X_3|S_3)P(S_3|S_2)P(X_2|S_2)P(S_2|S_1)P(X_1|S_1)P(S_1)=\boldsymbol\iota'\boldsymbol{F}_3\boldsymbol{P}'\boldsymbol{F}_2\boldsymbol{P}'\boldsymbol{F}_1\boldsymbol{P}'\boldsymbol{\pi}$$
I think it makes more sense to keep the product in that order, although you can take the transpose of the whole thing to get the article's expression. (Recall that $\boldsymbol{F}_t=\boldsymbol{F}_t'$.) Such minutiae aren't really too important. The big-picture takeaway is that this matrix product is not a product of scalar likelihoods for each $t$. It is sensitive to long-term correlations between the $S_t$, which supposedly can and should be ignored for the authors' purposes.
