Suppose we have a $\mathbb{K}$-vector-space $V$ which has finite dimension, so in particular $V=\langle v_1,\dots,v_n \rangle_{\mathbb{K}}$ ($V$ is generated as a vector space by $v_1,\dots,v_n$). Suppose $V \subseteq A$ where $A$ is a $\mathbb{K}$-algebra (commutative with $1$ and the scalar multiplication of $V$ is just the scalar multiplication of $A$). Then I claim that $$ \langle V \rangle_A=\langle v_1, \dots, v_n \rangle_A $$ where we write $\langle B \rangle_A$ for the $\mathbb{K}$-algebra (commutative with $1$) generated by a set $B \subseteq A$. In particular, a finite dimensional vector space generates a finitely generated algebra.
Proof idea: Elements in $\langle V \rangle_A $ are of the form $\sum_{i=1}^m \alpha_i \cdot x_1 \cdots x_{m_i} + \sum_{k=1}^{l} \beta_k \cdot 1$ for $\alpha_i, \beta_k \in \mathbb{K},x_j \in V$ (where $1$ is the identity in $A$). All elements $x_j$ can be written as $x_j = \sum_{i=1}^n \gamma_{j,i} \, v_i$ for $\gamma_{j,i} \in \mathbb{K}$. Hence, all elements in $\langle V \rangle_A$ can be written as polynomials in $v_1,\dots,v_n$ over $\mathbb{K}$.
Is the statement above true? If so, do you know of any references for it?
The statement is true and the proof you gave works just fine. Here's another idea that requires writing fewer sums. $\langle B \rangle_A$ is the smallest $\mathbb K$ subalgebra of $A$ that contains $B$. So certainly $\langle v_1, \dots v_n \rangle_A \subseteq \langle V\rangle_A$. On the other hand, let $v\in V$. Then $v=\sum a_i v_i$ so $v \in \langle v_1,\dots,v_n\rangle_A$. Hence, $V\subseteq \langle v_1,\dots,v_n\rangle_A$ and we have $\langle V\rangle_A \subseteq \langle v_1,\dots,v_n\rangle_A \subseteq \langle V\rangle_A$. This is the desired equality.