. . . Eq. (12.118)¹

Let's acquiesce to Griffiths' notation for this problem.

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. . . transformation,²

The Einstein summation convention is assumed: sum over all repeated indices.

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. . . direction.³

Eqs. (12.108) are for a "boost" in the $\Hat{x}$ direction. This is rather tedious, especially since the result is rather obvious when you're done, but everyone should do it once in order to understand that the definition of $F^{\mu\nu}$ is more useful formally (for elegantly expressing the essence of electromagnetism) than for solving practical problems.

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. . . respectively.⁴

In the moving frame, the plates appear to be moving in the opposite direction; thus the current in the positive plate is in the $-\Hat{x}$ direction and that in the negative plate is in the $+\Hat{x}$ direction.

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