Relationship between Taylor and Weierstrass theorem

Morally, these two theorems shouldn't say much about each other. If either were to imply the other, it seems less absurd that Weierstrass implies Taylor: if all we know is Taylor then we only have very weak data about a generic continuous function (if $f$ were to be absolutely continuous then we could perhaps use its antiderivative to some end, but generally speaking, no). But this direction also seems unlikely, because Weierstrass gives no control at all on the degree of the polynomial.

As rych points out, the two kinds of approximations suggested by these two theorems are different. Weierstrass' approximation is uniform which means that no point in the approximation can be further than a specified tolerance from the original. But Taylor's approximation is a much more subtle condition: no point $p(x)$ in the approximation can be further than $\varepsilon (x-a)^k$ from $f(x)$. This is a stronger condition: it means, for instance, that $p(a)=0$, but this need not be true at any point of a Weierstrass approximation.

Yes, they are both about "approximation". But approximating uniformly on an interval is quite different from approximating locally about the point. The two results also require functions of different classes: continuous vs. k-times differentiable.