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I have a confusion about the definition of flat sheaf of module. I feel very stupid about it and kind of shy to ask it here but at the same time I can not get it rid of my mind. So, I apologies in advance if my question turns out to be something stupid!

Let $f: X \rightarrow Y$ be a morphism of schemes and $\mathcal{F}$ be a sheaf of $\mathcal{O}_X$ module. Then $\mathcal{F}$ is flat over $Y$ at a point $x\in X$ if $\mathcal{F_x}$ is flat $\mathcal{O}_{y,Y}$-module \mathcal{O}_{y,Y}$ -module where $y=f(x)$ and $\mathcal{F_x}$ is considered as an $\mathcal{O_{y,Y}}$-module \mathcal{O}_{y,Y}$ -module by the natural map $f^#$: $\mathcal{O_{y,Y}}$ $\rightarrow f^{\sharp}:\mathcal{O}_{y,Y}\rightarrow \mathcal{O}_{x,X}$-module.mathcal{O}_{x,X}$.

Now, my question is, 'Is : is the above definition equivalent as saying $(f_*{\mathcal{F}})_{y}$ (f_{*} \mathcal{F})_{y}$ is flat as $\mathcal{O}{y,Y}$ module?' \mathcal{O}_{y,Y}$ module?

Similarly, 'Is is $\mathcal{F}$ flat at every point of $X$ over $Y$ is equivalent as saying $f_*\mathcal{F}$ is a flat $\mathcal{O}_Y$-module?'

This question came in my mind when I tried to prove the following result:

'Let $f:X\rightarrow Y$ be a finite morphism of noetherian schemes and $\mathcal{F}$ be a coherent sheaf on $X$. Then $\mathcal{F}$ is flat over $Y$ if and only if $f_*\mathcal{F}$ if locally free on $Y$.'

Let's assume $\mathcal{F}$ is flat first. Now, since $f$ is finite and $\mathcal{F}$ is coherent, therefore $f_*\mathcal{F}$ is coherent on $Y$.

At this point it appears to me that, if flatness of $\mathcal{F}$ implies that $f_*\mathcal{F}$ is flat over $Y$ then localizing at a point of $Y$ and then using the equivalence of free module and flat module over a noetherian local ring I can complete the proof.

Similarly, assuming $f_*\mathcal{F}$ is free on $Y$ implies it's flat on $Y$, then it seems to me that this should be same as saying $\mathcal{F}$ is flat over $Y$.

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About the Definition of Flat Morphism (Flat Sheaf)

I have a confusion about the definition of flat sheaf of module. I feel very stupid about it and kind of shy to ask it here but at the same time I can not get it rid of my mind. So, I apologies in advance if my question turns out to be something stupid!

Let $f: X \rightarrow Y$ be a morphism of schemes and $\mathcal{F}$ be a sheaf of $\mathcal{O}_X$ module. Then $\mathcal{F}$ is flat over $Y$ at a point $x\in X$ if $\mathcal{F_x}$ is flat $\mathcal{O}_{y,Y}$-module where $y=f(x)$ and $\mathcal{F_x}$ is considered as an $\mathcal{O_{y,Y}}$-module by the natural map $f^#$: $\mathcal{O_{y,Y}}$ $\rightarrow \mathcal{O}_{x,X}$-module.

Now, my question is, 'Is the above definition equivalent as saying $(f_*{\mathcal{F}})_{y}$ is flat as $\mathcal{O}{y,Y}$ module?'

Similarly, 'Is $\mathcal{F}$ flat at every point of $X$ over $Y$ is equivalent as saying $f_*\mathcal{F}$ is flat $\mathcal{O}_Y$-module?'

This question came in my mind when I tried to prove the following result:

'Let $f:X\rightarrow Y$ be a finite morphism of noetherian schemes and $\mathcal{F}$ be a coherent sheaf on $X$. Then $\mathcal{F}$ is flat over $Y$ if and only if $f_*\mathcal{F}$ if locally free on $Y$.'

Let's assume $\mathcal{F}$ is flat first. Now, since $f$ is finite and $\mathcal{F}$ is coherent, therefore $f_*\mathcal{F}$ is coherent on $Y$.

At this point it appears to me that, if flatness of $\mathcal{F}$ implies that $f_*\mathcal{F}$ is flat over $Y$ then localizing at a point of $Y$ and then using the equivalence of free module and flat module over a noetherian local ring I can complete the proof.

Similarly, assuming $f_*\mathcal{F}$ is free on $Y$ implies it's flat on $Y$, then it seems to me that this should be same as saying $\mathcal{F}$ is flat over $Y$.