Representation of a linear functional in vector space

Your assumption is that $\ker{\varphi} \supseteq \bigcap_{i=1}^k \ker{\varphi_i}$.

Consider the linear map $\ell \colon X \to \mathbb{R}^k$ given by $\ell(x) = (\varphi_1(x),\dots,\varphi_k(x))$ and let $V = \operatorname{im}\ell = \{\ell(x):x \in X\} \subseteq \mathbb{R}^k$ be the image. We have $\ker{\ell} = \bigcap_{i=1}^k \ker{\varphi_{i}} \subseteq \ker\varphi$. Therefore $\varphi = \tilde{\varphi} \circ \ell$ for some linear functional $\tilde{\varphi}\colon V \to \mathbb{R}$ [explicitly, $\tilde{\varphi}(v) = \varphi(x)$ where $x$ is such that $\ell(x) = v$. This is well-defined and linear.]

Every linear functional defined on a subspace $V$ of $\mathbb{R}^k$ can be extended to a linear functional on all of $\mathbb{R}^k$ (write $\mathbb{R}^k = V \oplus V^{\bot}$ and set the extension to be zero on $V^{\bot}$) and every linear functional on $\mathbb{R}^k$ is of the form $\psi(y) = \sum_{i=1}^k a_i y_i$. Thus, there are $\lambda_1,\dots,\lambda_k \in \mathbb{R}$ such that $\tilde\varphi(v) = \sum_{i=1}^k \lambda_i v_i$ for all $v \in V$. In other words, $\varphi = \sum_{i=1}^k \lambda_i \varphi_i$.

Look at Rudin's "Functional Analysis" Lemma 3.9. The only issue I see is that the proof requires the extension of a functional from a subspace of a finite dimensional space to the entire finite dimensional space, but this is purely algebraic as far as I can see.