Lorentz Transformation and Lorentz Invariance
An arbitrary Lorentz transformation, including rotations, can be written as a $4$-by-$4$ tensor $\Lambda^\mu_\nu$ satisfying the following:
\begin{equation}
\Lambda^\rho_\mu \eta_{\rho\sigma} \Lambda^\sigma_\nu = \eta_{\mu\nu}
\end{equation}
Such a Lorentz transformation acts on a $4$-vector $u$ as:
\begin{equation}
u'^\mu = \Lambda^\mu_\nu u^\nu
\end{equation}
to give a new $4$-vector $u'$.
This ensures the Lorentz invariance of scalar quantities such as the Lagrangian. To be more specific, under any Lorentz transformation,
\begin{align}
L \rightarrow L' &= \frac{m}{2}\eta_{\mu\nu} \left(\Lambda^\mu_\rho u^\rho\right) \left(\Lambda^\nu_\sigma u^\sigma\right) \\
&= \frac{m}{2} \left(\Lambda^\mu_\rho \eta_{\mu\nu} \Lambda^\nu_\sigma\right) u^\rho u^\sigma \\
&= \frac{m}{2} \eta_{\rho\sigma} u^\rho u^\sigma
\end{align}
Therefore, the Lagrangian for a free massive particle is Lorentz invariant.
Special-relativistic Equation of Motion
So, what does such a Lagrangian tell us? What is the dynamics of a free massive particle?
In special relativity, a point particle's position in the Minkowski spacetime (space of coordinate $4$-vectors) is parametrized by its proper time $\tau$. In other words, the $4$-coordinate of the particle is a $4$-vector whose each component is a function of a single real variable $\tau$. Naturally, the $4$-velocity $u^\mu = \dot{x}^\mu$ is also a $4$-vector whose components are only explicitly functions of the proper time $\tau$.
This is different from what happens in classical dynamics, in which a point particle's position and velocity are $3$-dimensional Euclidean vectors parametrized by the variable $t$, the regular time.
Hence, the Euler-Lagrange equation naturally gets modified from:
\begin{equation}
\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial L}{\partial \dot{x}^i}\right) - \frac{\partial L}{\partial x^i} = 0
\end{equation}
to
\begin{equation}
\frac{\mathrm{d}}{\mathrm{d}\tau}\left(\frac{\partial L}{\partial \dot{x}^\mu}\right) - \frac{\partial L}{\partial x^\mu} = 0
\end{equation}
Note that the 'dot' in special relativity implies a derivative with respect to the proper time $\tau$. Hence, $\dot{x}^0 = \dot{t}$ is not trivially $1$, for instance.
Now, we plug in the free special-relativistic point-particle Lagrangian into the new Euler-Lagrange equation to obtain the special-relativistic equation of motion. Note that $u^\mu = \dot{x}^\mu$.
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}\tau}\left(\frac{\partial L}{\partial \dot{x}^\mu}\right) - \frac{\partial L}{\partial x^\mu} &=
\frac{\mathrm{d}}{\mathrm{d}\tau}\left(\frac{\partial}{\partial \dot{x}^\mu} \left(\frac{m}{2}\eta_{\mu\nu}u^\mu u^\nu\right)\right) - \frac{\partial}{\partial x^\mu} \left(\frac{m}{2}\eta_{\mu\nu}u^\mu u^\nu\right) \\
&= \frac{\mathrm{d}}{\mathrm{d}\tau}\left(\frac{\partial}{\partial \dot{x}^\mu} \left(\frac{m}{2}\eta_{\mu\nu}u^\mu u^\nu\right)\right)
\end{align}
since the second-term is zero, since the free-particle Lagrangian only depends on the velocity $u$ and no the (spacetime) position $x$.
Now, to address the last question in the OP and thereby compute the remaining term, it is useful to consider an analogous expression in the 3-dimensional Euclidean space and write it in a vector form, not using the index notation. That is,
\begin{align}
\vec{\nabla} x^2 &= \vec{\nabla} \left(\vec{x} \cdot \vec{x} \right) \\
&= \left(\vec{x} \cdot \vec{\nabla} \right) \vec{x} + \left(\vec{x} \cdot \vec{\nabla} \right) \vec{x} + \vec{x} \times \left(\vec{\nabla} \times \vec{x}\right) + \vec{x} \times \left(\vec{\nabla} \times \vec{x}\right) \\
&= 2\vec{x}
\end{align}
Similarly, we have here that:
\begin{align}
\frac{\partial u^\mu u^\nu}{\partial \dot{x}^\nu}
&= \frac{\partial u^\mu}{\partial \dot{x}^\nu} u^\nu
+ u^\mu \frac{\mathrm{d} u^\nu}{\mathrm{d} \dot{x}^\nu} \\
&= \delta^\mu_\nu u^\nu + u^\mu \cdot 1 \\
&= 2u^\mu
\end{align}
Therefore, the Euler-Lagrange equation reduces to:
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}\tau}\left(\frac{\partial L}{\partial \dot{x}^\mu}\right) - \frac{\partial L}{\partial x^\mu} &=
\frac{\mathrm{d}}{\mathrm{d}\tau}\left(\frac{\partial}{\partial \dot{x}^\mu}\left(\frac{m}{2}\eta_{\mu\nu}u^\mu u^\nu\right)\right) \\
&= \frac{\mathrm{d}}{\mathrm{d}\tau}\left(\frac{m}{2}\eta_{\mu\nu} \left(2u^\nu\right)\right) \\
&= \frac{\mathrm{d}}{\mathrm{d}\tau}\left(m\eta_{\mu\nu}u^{\nu}\right) \\
&= 0
\end{align}
Because the Lagrangian itself is Lorentz invariant, the equation of motion is, of course, Lorentz invariant.
