From Algebra: Chapter $0$ by Aluffi:
I'm pretty sure $$\varphi(\sum m_{i_1 \cdots i_n} x_1^{i_1} \cdots x_n^{i_n})= \sum \alpha(m_{i_1 \cdots i_n}) f(1)^{i_1} \cdots f(n)^{i_n}.$$
So, we have \begin{align*} \varphi(r\sum m_{i_1 \cdots i_n} x_1^{i_1} \cdots x_n^{i_n}) &= \varphi(\sum rm_{i_1 \cdots i_n} x_1^{i_1} \cdots x_n^{i_n})\\ &=\sum \alpha(rm_{i_1 \cdots i_n}) f(1)^{i_1} \cdots f(n)^{i_n}\\ &= \sum \alpha(r)\alpha(m_{i_1 \cdots i_n}) f(1)^{i_1} \cdots f(n)^{i_n}\\ &=\alpha(r) \sum \alpha(m_{i_1 \cdots i_n}) f(1)^{i_1} \cdots f(n)^{i_n}\\ &=\alpha(r) \varphi(\sum m_{i_1 \cdots i_n} x_1^{i_1} \cdots x_n^{i_n}) \end{align*}
How do we show $$\varphi(r\sum m_{i_1 \cdots i_n} x_1^{i_1} \cdots x_n^{i_n})= r\varphi(\sum m_{i_1 \cdots i_n} x_1^{i_1} \cdots x_n^{i_n})?$$
How is this automatic?
Your 'pretty sure' line proves it - which is indeed true -, using that $\alpha$ is a ring homomorphism: $$\varphi(rmx_1^{i_1}\dots x_n^{i_n}) =\alpha(rm)\varphi(x_1^{i_1}\dots x_n^{i_n}) =\alpha(r)\cdot\alpha(m) \varphi(x_1^{i_1}\dots x_n^{i_n})=\alpha(r)\cdot \varphi(m\, x_1^{i_1}\dots x_n^{i_n}) $$