If you were to place an electric field sensor fixed at a point in space – conceptually, let's say a small positively charged fin that registered a force upward or downward in proportion to the electric field value and direction (up or down) at that point – and then had a plane EM wave pass by horizontally (in the $x$ direction), you would see this sensor register a sinusoidal force over time
$$\vec F=F_0\cos(\omega t) \hat y$$
If we say the sensor has unit charge, the corresponding $\bf E$ field at that point is
$$\vec E=E_0\cos(\omega t) \hat y$$
Which oscillates in time between $+E_0$ and $-E_0$ in a sinusoidal fashion.
In addition, if while the plane wave is propagating in the $x$ direction, you froze time for an instant and examined the spatial variation of the $\bf E$ field by sliding your sensor in the $x$ direction, you would find the spatial distribution
$$\vec E=E_0\cos(kx) \hat y$$
This spatial sinusoidal "shape," when you unfreeze time, marches forward in the $x$ direction at the speed of light, in such a way that any point will have the time $\vec E(t)$ variation mentioned.
Thus a propagating EM plane wave is represented by:
$$\vec E=E_0\cos(\omega t-kx) \hat y$$
Where $\omega$ is the $2\pi×$ the frequency of the wave in Hertz, and $k$ is the wave number ($\rm cm^{-1}$), or 1 over the wavelength (in cmcm). It needs both terms inside the cosine to show the spatial distribution, and also that that distribution moves forward over time.
To your precise question, while the wave only moves linearly in the $x$ direction, the electric field points in the $+$ and $-y$ direction in a sinusoidal fashion. This is why it is a transverse wave.
A "plane wave" simply means that if you were to translate your sensor to any point on the $(y,z)$ plane at a fixed $x$ location, your $E$ field readings above would be exactly the same (at a given moment in time).
(Image Credit: https://www4.uwsp.edu/physastr/kmenning/Phys202/Lect16.html)