The ML inequality property in complex integral says $|\int_{c}f(z)dz| \leq ML$.
If I have two function in the integral, I can write the inequality: $|\int_{c}f(z)g(z)dz| \leq ML|\int_{c}g(z)dz| $ ?. It's that legal? $f(z)$ and $g(z)$ belongs to $L^2$
The estimate I know for contour integrals says that if $f$ is continuous on some path $\gamma:[a,b] \rightarrow \mathbb{C}$, we have $$\bigg|\int_{\gamma} f(z)\,dz \bigg| \leq M \cdot l(\gamma).$$ Here, $l(\gamma)$ is the length of the path $\gamma$, and $M = \max\{|f(z)|: z \in \gamma\}.$ When I say $z \in \gamma$, I'm abusing notation and I mean $z$ lies on the image of $\gamma$.
The inequality you propose in you question is not true in general, even if we suppose that in addition $f$ and $g$ square integrable. Indeed, let $\gamma (t) = e^{2\pi i t}$, which is just the unit circle. Then $f(z) = z$ and $g(z) = 1/z^2$ are continuous and square integrable along this path, and its a simple computation to verify that $$\int_{\gamma} f(z) g(z) dz = \int_{\gamma} \frac{1}{z} dz = 2\pi i.$$ On the other hand, $g(z)$ has a primitive in $\mathbb{C} \setminus \{0\}$, so $$\int_\gamma g(z) dz = \int_{\gamma} \frac{1}{z^2}dz = 0.$$ You can use these observations to conclude the proposed inequality is false in general.
However, you can can get a similar estimate to the one you proposed that may still be useful by applying the Cauchy-Schwarz inequality. If $f$ and $g$ are square integrable, and $M = \max\{|f(z)|: z \in \gamma\},$ we have $$\bigg|\int_{\gamma} f(z) \overline g(z) dz\bigg|^2 \leq \int_{\gamma} |f(z)|^2\,dz \cdot \int_{\gamma} |g(z)|^2 \,dz \leq M^2 \cdot l(\gamma) \cdot \int_{\gamma} |g(z)|^2\,dz.$$