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Let $a:=\mu_1(0)$ and $b:=\mu_n(1)$. The linear span of the $\{f_k\}_{k\in\mathbb{N}}$ is $\big\{\sum_{j=1}^nP\circ\mu_j : P\in\mathbb{R}[x]\big\}$, whose closure in $C^0([0,1]),\|\cdot\|_\infty$ contains the space $\big\{\sum_{j=1}^nf\circ\mu_j : f\in C^0([a,b])\big\}$, just because polynomials are uniformly dense in $C^0([a,b]) $ and $f\mapsto f\circ\mu_j$ are (linear) continuous maps. So the question is: can any $\alpha\in C^0([0,1])$ be written in the form $$\alpha=\sum_{j=1}^nf\circ\mu_j$$ for some $f\in C^0([a,b]) $?

Let's define inductively a finite sequence in $[a,b]$ putting $c_0:=\mu_n(0)$ and $c_{k+1}:=\mu_n(\mu_{n-1}^{-1}(c_k))$ until we reach some $c_K$ out of the range of $\mu_{n-1}$, which happens in finitely many steps, because $$c_{k+1}-c_k=\mu_n(\mu_{n-1}^{-1}(c_k))-\mu_{n-1}(\mu_{n-1}^{-1}(c_k))\ge\delta:=\min_{0\ge t\ge1}\mu_n(t)-\mu_{n-1}(t)>0.$$ Since $\mu_n:[0,1]\to[c_0,b]$ is invertible, we can define arbitrarily $f$ on $[a,c_0]$, with the condition $$\alpha(0)=\sum_{j=1}^nf(\mu_j(0))$$ and state the functional equation for $f$ on $[c_0,b]$ equivalently as: $$f(y) =\alpha(\mu_n^{-1}(y))-\sum_{j=1}^{n-1}f(\mu_j(\mu_n^{-1}(y)),\qquad y\in[c_0,b].$$ But this equation is self-solving: the RHS gives the unique extension of $f$ to the interval $[a,c_{1}]$, then to $[a,c_{2}]$, till we cover $[a,b]$.

edit If you change the definition of $f_k$, keeping the assumptions on the $\mu_k$, to $$f_k:=[a^k]{1\over(1-a\mu_1)(1-a\mu_2)\dots(1-a\mu_n)}$$ that is $$f_k=\sum_{j=1}^n\lambda_j\mu_j^k$$ with $$\lambda_j(t):=\prod_{1\le i\le n\atop i\ne j}{\mu_j(t)\over\mu_i(t)-\mu_j(t)},$$ Then the linear span of the $\{f_k\}_{k\in\mathbb{N}}$ is $\big\{\sum_{j=1}^n\lambda_j(P\circ\mu_j) : P\in\mathbb{R}[x]\big\}$, and the closure of this space in $C^0([0,1]),\|\cdot\|_\infty$ contains the space $\big\{\sum_{j=1}^n\lambda_j(f\circ\mu_j) : f\in C^0([a,b])\big\}$. To show the latter space is the whole space $C^0([0,1])$, means solvability for $f\in C^0([a,b]) $ of the functional equation $$\alpha(t)=\sum_{j=1}^n \lambda_j(t)f(\mu_j(t)),\quad t\in[0,1]$$ for any datum $\alpha\in C^0([0,1])$. To this end we can argue as before: choose a continuous $f$ on $[a,c_0]$ satisfying the functional equation at $x=0$: $$\alpha(0)=\sum_{j=1}^n \lambda_j(0)f(\mu_j(0))$$

And extend it to a solution $f\in C^0([a,b])$ now iterating for $j=1,2\dots$

$$f(y) ={\alpha(\mu_n^{-1}(y))\over\lambda_n(\mu_n^{-1}(y))}-\sum_{j=1}^{n-1}{\lambda_j(\mu_n^{-1}(y))\over\lambda_n(\mu_n^{-1}(y))}f(\mu_j(\mu_n^{-1}(y)),\qquad y\in[a,c_j].$$

Let $a:=\mu_1(0)$ and $b:=\mu_n(1)$. The linear span of the $\{f_k\}_{k\in\mathbb{N}}$ is $\big\{\sum_{j=1}^nP\circ\mu_j : P\in\mathbb{R}[x]\big\}$, whose closure in $C^0([0,1]),\|\cdot\|_\infty$ contains the space $\big\{\sum_{j=1}^nf\circ\mu_j : f\in C^0([a,b])\big\}$, just because polynomials are uniformly dense in $C^0([a,b]) $ and $f\mapsto f\circ\mu_j$ are (linear) continuous maps. So the question is: can any $\alpha\in C^0([0,1])$ be written in the form $$\alpha=\sum_{j=1}^nf\circ\mu_j$$ for some $f\in C^0([a,b]) $?

Let's define inductively a finite sequence in $[a,b]$ putting $c_0:=\mu_n(0)$ and $c_{k+1}:=\mu_n(\mu_{n-1}^{-1}(c_k))$ until we reach some $c_K$ out of the range of $\mu_{n-1}$, which happens in finitely many steps, because $$c_{k+1}-c_k=\mu_n(\mu_{n-1}^{-1}(c_k))-\mu_{n-1}(\mu_{n-1}^{-1}(c_k))\ge\delta:=\min_{0\ge t\ge1}\mu_n(t)-\mu_{n-1}(t)>0.$$ Since $\mu_n:[0,1]\to[c_0,b]$ is invertible, we can define arbitrarily $f$ on $[a,c_0]$, with the condition $$\alpha(0)=\sum_{j=1}^nf(\mu_j(0))$$ and state the functional equation for $f$ on $[c_0,b]$ equivalently as: $$f(y) =\alpha(\mu_n^{-1}(y))-\sum_{j=1}^{n-1}f(\mu_j(\mu_n^{-1}(y)),\qquad y\in[c_0,b].$$ But this equation is self-solving: the RHS gives the unique extension of $f$ to the interval $[a,c_{1}]$, then to $[a,c_{2}]$, till we cover $[a,b]$.

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edit. If you change the definition of $f_k$, keeping the assumptions on the $\mu_k$, and also assuming $\mu_n>0$, to $$f_k:=[a^k]{1\over(1-a\mu_1)(1-a\mu_2)\dots(1-a\mu_n)}$$ that is $$f_k=\sum_{j=1}^n\lambda_j\mu_j^k$$ with $$\lambda_j(t):=\prod_{1\le i\le n\atop i\ne j}{\mu_j(t)\over \mu_j(t)-\mu_i(t)},$$ Then the linear span of the $\{f_k\}_{k\in\mathbb{N}}$ is $\big\{\sum_{j=1}^n\lambda_j(P\circ\mu_j) : P\in\mathbb{R}[x]\big\}$, and the closure of this space in $C^0([0,1]),\|\cdot\|_\infty$ contains the space $\big\{\sum_{j=1}^n\lambda_j(f\circ\mu_j) : f\in C^0([a,b])\big\}$. To show the latter space is the whole space $C^0([0,1])$, means solvability for $f\in C^0([a,b]) $ of the functional equation $$\alpha(t)=\sum_{j=1}^n \lambda_j(t)f(\mu_j(t)),\quad t\in[0,1]$$ for any datum $\alpha\in C^0([0,1])$. To this end we can argue as before: choose a continuous $f$ on $[a,c_0]$ satisfying the functional equation at $x=0$: $$\alpha(0)=\sum_{j=1}^n \lambda_j(0)f(\mu_j(0))$$

And extend it to a solution $f\in C^0([a,b])$ now iterating for $j=1,2\dots$

$$f(y) ={\alpha(\mu_n^{-1}(y))\over\lambda_n(\mu_n^{-1}(y))}-\sum_{j=1}^{n-1}{\lambda_j(\mu_n^{-1}(y))\over\lambda_n(\mu_n^{-1}(y))}f(\mu_j(\mu_n^{-1}(y)),\qquad y\in[a,c_j].$$

Let $a:=\mu_1(0)$ and $b:=\mu_n(1)$. The linear span of the $\{f_k\}_{k\in\mathbb{N}}$ is $\big\{\sum_{j=1}^nP\circ\mu_j : P\in\mathbb{R}[x]\big\}$, whose closure in $C^0([0,1]),\|\cdot\|_\infty$ contains the space $\big\{\sum_{j=1}^nf\circ\mu_j : f\in C^0([a,b])\big\}$, just because polynomials are uniformly dense in $C^0([a,b]) $ and $f\mapsto f\circ\mu_j$ are (linear) continuous maps. So the question is: can any $\alpha\in C^0([0,1])$ be written in the form $$\alpha=\sum_{j=1}^nf\circ\mu_j$$ for some $f\in C^0([a,b]) $?

Let's define inductively a finite sequence in $[a,b]$ putting $c_0:=\mu_n(0)$ and $c_{k+1}:=\mu_n(\mu_{n-1}^{-1}(c_k))$ until we reach some $c_K$ out of the range of $\mu_{n-1}$, which happens in finitely many steps, because $$c_{k+1}-c_k=\mu_n(\mu_{n-1}^{-1}(c_k))-\mu_{n-1}(\mu_{n-1}^{-1}(c_k))\ge\delta:=\min_{0\ge x\ge1}\mu_n(x)-\mu_{n-1}(x)>0.$$ Since $\mu_n:[0,1]\to[c_0,b]$ is invertible, we can define arbitrarily $f$ on $[a,c_0]$, with the condition $$\alpha(0)=\sum_{j=1}^nf(\mu_j(0))$$ and state the functional equation for $f$ on $[c_0,b]$ equivalently as: $$f(y) =\alpha(\mu_n^{-1}(y))-\sum_{j=1}^{n-1}f(\mu_j(\mu_n^{-1}(y)),\qquad y\in[c_0,b].$$ But this equation is self-solving: the RHS gives the unique extension of $f$ to the interval $[a,c_{1}]$, then to $[a,c_{2}]$, till we cover $[a,b]$.

Let $a:=\mu_1(0)$ and $b:=\mu_n(1)$. The linear span of the $\{f_k\}_{k\in\mathbb{N}}$ is $\big\{\sum_{j=1}^nP\circ\mu_j : P\in\mathbb{R}[x]\big\}$, whose closure in $C^0([0,1]),\|\cdot\|_\infty$ contains the space $\big\{\sum_{j=1}^nf\circ\mu_j : f\in C^0([a,b])\big\}$, just because polynomials are uniformly dense in $C^0([a,b]) $ and $f\mapsto f\circ\mu_j$ are (linear) continuous maps. So the question is: can any $\alpha\in C^0([0,1])$ be written in the form $$\alpha=\sum_{j=1}^nf\circ\mu_j$$ for some $f\in C^0([a,b]) $?

Let's define inductively a finite sequence in $[a,b]$ putting $c_0:=\mu_n(0)$ and $c_{k+1}:=\mu_n(\mu_{n-1}^{-1}(c_k))$ until we reach some $c_K$ out of the range of $\mu_{n-1}$, which happens in finitely many steps, because $$c_{k+1}-c_k=\mu_n(\mu_{n-1}^{-1}(c_k))-\mu_{n-1}(\mu_{n-1}^{-1}(c_k))\ge\delta:=\min_{0\ge t\ge1}\mu_n(t)-\mu_{n-1}(t)>0.$$ Since $\mu_n:[0,1]\to[c_0,b]$ is invertible, we can define arbitrarily $f$ on $[a,c_0]$, with the condition $$\alpha(0)=\sum_{j=1}^nf(\mu_j(0))$$ and state the functional equation for $f$ on $[c_0,b]$ equivalently as: $$f(y) =\alpha(\mu_n^{-1}(y))-\sum_{j=1}^{n-1}f(\mu_j(\mu_n^{-1}(y)),\qquad y\in[c_0,b].$$ But this equation is self-solving: the RHS gives the unique extension of $f$ to the interval $[a,c_{1}]$, then to $[a,c_{2}]$, till we cover $[a,b]$.

edit If you change the definition of $f_k$, keeping the assumptions on the $\mu_k$, to $$f_k:=[a^k]{1\over(1-a\mu_1)(1-a\mu_2)\dots(1-a\mu_n)}$$ that is $$f_k=\sum_{j=1}^n\lambda_j\mu_j^k$$ with $$\lambda_j(t):=\prod_{1\le i\le n\atop i\ne j}{\mu_j(t)\over\mu_i(t)-\mu_j(t)},$$ Then the linear span of the $\{f_k\}_{k\in\mathbb{N}}$ is $\big\{\sum_{j=1}^n\lambda_j(P\circ\mu_j) : P\in\mathbb{R}[x]\big\}$, and the closure of this space in $C^0([0,1]),\|\cdot\|_\infty$ contains the space $\big\{\sum_{j=1}^n\lambda_j(f\circ\mu_j) : f\in C^0([a,b])\big\}$. To show the latter space is the whole space $C^0([0,1])$, means solvability for $f\in C^0([a,b]) $ of the functional equation $$\alpha(t)=\sum_{j=1}^n \lambda_j(t)f(\mu_j(t)),\quad t\in[0,1]$$ for any datum $\alpha\in C^0([0,1])$. To this end we can argue as before: choose a continuous $f$ on $[a,c_0]$ satisfying the functional equation at $x=0$: $$\alpha(0)=\sum_{j=1}^n \lambda_j(0)f(\mu_j(0))$$

And extend it to a solution $f\in C^0([a,b])$ now iterating for $j=1,2\dots$

$$f(y) ={\alpha(\mu_n^{-1}(y))\over\lambda_n(\mu_n^{-1}(y))}-\sum_{j=1}^{n-1}{\lambda_j(\mu_n^{-1}(y))\over\lambda_n(\mu_n^{-1}(y))}f(\mu_j(\mu_n^{-1}(y)),\qquad y\in[a,c_j].$$

Let $a:=\mu_1(0)$ and $b:=\mu_n(1)$. The linear span of the $\{f_k\}_{k\in\mathbb{N}}$ is $\big\{\sum_{j=1}^nP\circ\mu_j : P\in\mathbb{R}[x]\big\}$, whose closure in $C^0([0,1]),\|\cdot\|_\infty$ contains the space $\big\{\sum_{j=1}^nf\circ\mu_j : f\in C^0([a,b])\big\}$, just because polynomials are uniformly dense in $C^0([a,b]) $ and $f\mapsto f\circ\mu_j$ are (linear) continuous maps. So the question is: can any $\alpha\in C^0([0,1])$ be written in the form $$\alpha=\sum_{j=1}^nf\circ\mu_j$$ for some $f\in C^0([a,b]) $?

Let's define inductively a finite sequence in $[a,b]$ putting $c_0:=\mu_n(0)$ and $c_{k+1}:=\mu_n(\mu_{n-1}^{-1}(c_k))$ until we reach some $c_K$ out of the range of $\mu_{n-1}$, which happens in finitely many steps, because $$c_{k+1}-c_k=\mu_n(\mu_{n-1}^{-1}(c_k))-\mu_{n-1}(\mu_{n-1}^{-1}(c_k))\ge\delta:=\min_{0\ge x\ge1}\mu_n(x)-\mu_{n-1}(x)>0.$$ Since $\mu_n:[0,1]\to[c_0,b]$ is invertible, we can define arbitrarily $f$ on $[a,c_0]$, and state the functional equation for $f$ on $[c_0,b]$ equivalently as: $$f(y) =\alpha(\mu_n^{-1}(y))-\sum_{j=1}^{n-1}f(\mu_j(\mu_n^{-1}(y)),\qquad y\in[c_0,b].$$ But this equation is self-solving: the RHS gives the unique extension of $f$ to the interval $[a,c_{1}]$, then to $[a,c_{2}]$, till we cover $[a,b]$.

Let $a:=\mu_1(0)$ and $b:=\mu_n(1)$. The linear span of the $\{f_k\}_{k\in\mathbb{N}}$ is $\big\{\sum_{j=1}^nP\circ\mu_j : P\in\mathbb{R}[x]\big\}$, whose closure in $C^0([0,1]),\|\cdot\|_\infty$ contains the space $\big\{\sum_{j=1}^nf\circ\mu_j : f\in C^0([a,b])\big\}$, just because polynomials are uniformly dense in $C^0([a,b]) $ and $f\mapsto f\circ\mu_j$ are (linear) continuous maps. So the question is: can any $\alpha\in C^0([0,1])$ be written in the form $$\alpha=\sum_{j=1}^nf\circ\mu_j$$ for some $f\in C^0([a,b]) $?

Let's define inductively a finite sequence in $[a,b]$ putting $c_0:=\mu_n(0)$ and $c_{k+1}:=\mu_n(\mu_{n-1}^{-1}(c_k))$ until we reach some $c_K$ out of the range of $\mu_{n-1}$, which happens in finitely many steps, because $$c_{k+1}-c_k=\mu_n(\mu_{n-1}^{-1}(c_k))-\mu_{n-1}(\mu_{n-1}^{-1}(c_k))\ge\delta:=\min_{0\ge x\ge1}\mu_n(x)-\mu_{n-1}(x)>0.$$ Since $\mu_n:[0,1]\to[c_0,b]$ is invertible, we can define arbitrarily $f$ on $[a,c_0]$, with the condition $$\alpha(0)=\sum_{j=1}^nf(\mu_j(0))$$ and state the functional equation for $f$ on $[c_0,b]$ equivalently as: $$f(y) =\alpha(\mu_n^{-1}(y))-\sum_{j=1}^{n-1}f(\mu_j(\mu_n^{-1}(y)),\qquad y\in[c_0,b].$$ But this equation is self-solving: the RHS gives the unique extension of $f$ to the interval $[a,c_{1}]$, then to $[a,c_{2}]$, till we cover $[a,b]$.