Chris's note is in the correct direction, but I've had a few questions about this issue, so I think I'll give a full solution. Firstly, note what we're trying to show: $$\mathbb{E}[f(W_t) |\mathcal{F}^W_s]=\mathbb{E}[f(W_t) |W_s],$$ for all \(0\leq s \leq t\) and for all functions \(f(\cdot)\). This is the definition of the Markov property.

Let's use the independence lemma to express the left-hand side. One should rewrite \(f\) so that it takes two arguments, and then apply the independence lemma:$$\mathbb{E}[f(W_t) |\mathcal{F}^W_s]=\mathbb{E}[f(W_s,W_t-W_s) |\mathcal{F}^W_s]=g(W_s),$$

where \(g(x)=\mathbb{E}[f(x,W_t-W_t)]\). In the first step, I am essentially defining a new function that takes two arguments, adds them together, and then applies the old function (strictly speaking, this new function should have a new name, but it is conventional to retain the old symbol \(f\) because the function ends up doing the same thing - you could give a new name if you prefer: \( h(x,y)=f(x+y) \) ). The correct definition of this new function justifies the first step. The second step is justified by the independence lemma - it is important that you understand that it is applicable and that this is the correct application.

Now let's consider the right-hand side. Again we can write $$\mathbb{E}[f(W_t) |W_s]=\mathbb{E}[f(W_s,W_t-W_s) |W_s]=g(W_s),$$

where \(g\) is defined as above. Once these two steps are justified, the result is proved. The first step relies on the same definition of the modified version of \(f\). The second step uses the independence lemma. The crucial insight is that using the independence lemma in this slight different circumstance still gives the same form for unconditional expectation (which we have been calling \(g\)).

If you cannot follow this argument, I think it will be worthwhile to try to figure out exactly where you are struggling and to address those points.