Search for the word "intuitive" here. I do not understand how
$$E[X]=0\cdot \bar{a}+\frac{E[X]}{a}\cdot a$$
shows anything at all. I would like to see an intuitive proof of an upper bound of $P(x\geq a)$ which should be $$P(x\geq a)\leq \frac{E[X]}{a}.$$
For an increasing $\varphi$, $X\geq a \implies \varphi(X)\geq\varphi(a)$. Thus you can construct a similar argument. However, remember it is important to have $X$ a non-negative random variable in the original argument and thus if we are to replicate the argument the image of $\varphi$ should be non-negative.
Note that what you request, i.e.
$$P(|X|\geq a)\leq \frac{E[\varphi(X)]}{\varphi(a)}$$
can not be true in general. Take $X\sim N(0,1)$, $a=1$ and $\varphi$ the identity.