Suppose I have sequences $(f_{i,n})_n$ for $i=0,1,..,m$ of $M$-Lipschitz functions from $\mathbb{R}$ to $\mathbb{R}$. Each of those uniformly converges to an $M$-Lipschitz function $f_i$. Now consider $F_n:\mathbb{R} \to \mathbb{R}^k$ defined as $F_n(x)=\sum\limits_{i=0}^mv_if_{i,n}$ where $v_i \in \mathbb{R}^k$ that converges uniformly to a function $F$.
Is $F$ Lipschitz-continuous?
Yes. Notice that, for every $i = 0, \dots, m$ and $n\in\mathbb{N}$,
\begin{align} |f_{i,n}(x) - f_{i,n}(y)| &= |f_{i,n}(x)-f_{i}(x)| + |f_{i,n}(y)-f_{i}(y)| + |f_i(x)-f_i(y)|\\ &\leq 2\|f_{i,n}-f_{i}\|_{\infty} + M|x-y|. \end{align}
Thus,
\begin{align} |F(x)-F(y)| &= |F(x)-F_n(x)| + |F(y)-F_n(y)| + |F_n(x)-F_n(y)|\\ &\leq 2\|F-F_n\|_{\infty} + \sum_{i=0}^m |v_i| |f_{i,n}(x) - f_{i,n}(y)|\\ &\leq 2\|F-F_n\|_{\infty} + \sum_{i=0}^m |v_i| \Big(2\|f_{i,n}-f_{i}\|_{\infty} + M|x-y|\Big)\\ \end{align}
By uniform convergence, we have that $\|F-F_n\|_{\infty}\xrightarrow{n\to\infty} 0$ and $\|f_{i,n} - f_i\|_{\infty} \xrightarrow{n\to\infty} 0$ for each $i$. Passing the above inequality to the limit we find that $F$ is Lipschitz.
Edit: Notice that I haven't used the fact that the $f_{i,n}$ here are $M$-Lipschitz. Using this one can prove the second inequality directly:
\begin{align} |F(x)-F(y)| &\leq 2\|F-F_n\|_{\infty} + \sum_{i=0}^m |v_i| |f_{i,n}(x) - f_{i,n}(y)|\\ &\leq 2\|F-F_n\|_{\infty} + \sum_{i=0}^m |v_i| M|x-y|\\ \end{align}
and then it suffices to use the uniform convergence of the $F_n$.