Given an element $m$ in a coalgebra $C$, there always exists a finite-dimensional subcoalgebra $D \subset C$ containing $m$; this is the fundamental theorem for coalgebras. This obviously isn't the case for algebras. However, to make a proper analogue of this theorem for algebras, one should formulate this theorem using arrows and then dualize. An element of $C$ determines a linear map $F:\mathbb{k} \to C$ and vice versa. The fundamental theorem then becomes:
For any linear map $F: \mathbb{k} \to C$ there exists a map $G:\mathbb{k} \to D$, such that $i \circ G=F$, for $D$ some finite dimensional coalgebra, and $i$ an injective map of coalgebras.
Dualizing to algebras, one easily sees that the statement becomes equivalent to the question: is any linear map $f:A \to \mathbb{k}$ zero on an ideal of finite codimension, where $A$ is any algebra. Spelling this out for $k[x]$, one is led to the question: do there exist sequences $(f_{n})_{n}$ that do not satisfy any recursion equation. Does anyone know about a sequence like this, or an existence proof?
A perhaps less sensical question: does the reformulation of the fundamental theorem in terms of algebras have some nice geometrical interpretation? Say in the case of affine algebras?
Very interesting. Unfortunately the dual statement you suggest is false. Since $k$ is a field, $k [x]$ is a principal ideal domain. In particular, any non-trivial quotient of $k [x]$ is automatically finite-dimensional. Now, any linear map $f : k [x] \to k$ is freely and uniquely determined by the sequence $f(1), f(x), f(x^2), \ldots$, and $f$ factors through the quotient $k [x] / (p(x))$ only if $$a_0 f(x^n) + a_1 f(x^{n+1}) + \cdots + a_d f(x^{n+d}) = 0$$ for all $n$, where $$p(x) = a_0 + a_1 x + \cdots + a_d x^d$$ so to find an $f$ that does not factor through any non-trivial quotient of $k [x]$, it is enough to find a sequence that does not satisfy any linear recurrence equation (with constant coefficients). But a sequence that satisfies a linear recurrence equation grows at worst exponentially, so we can just take a super-exponential sequence, say $n!$.