Consider a Lagrangian density
$$\mathcal{L}(\phi, \nabla \phi) = \frac{1}{2} \, g^{\mu \nu} \, \partial_{\mu} \phi \; \partial_{\nu} \phi + V(\phi) \tag{1}$$
The equation of motion (EOM), i.e. the Euler-Lagrange equation, is
$$\nabla^{\mu} \nabla_{\mu} \phi - \frac{\partial V}{\partial \phi} = 0 \tag{2}$$
Assume now that $\phi = (\phi_0 + \varepsilon)$ and that $\phi_0$ satisfies equation (2). Inserting into said equation (2) and linearising yields
$$\nabla^{\mu} \nabla_{\mu} \varepsilon - \varepsilon \cdot \frac{\partial^2 V}{\partial \phi^2} \, \bigg|_{\phi_0} = 0 \tag{3}$$
The corresponding Lagrangian density that yields this EOM in $\varepsilon$ is
$$\tilde{\mathcal{L}}(\varepsilon, \nabla \varepsilon) = \frac{1}{2} \, g^{\mu \nu} \, \partial_{\mu} \varepsilon \; \partial_{\nu} \varepsilon + \varepsilon^2 \cdot \frac{1}{2} \, \frac{\partial^2 V}{\partial \phi^2} \, \bigg|_{\phi_0} \tag{4} $$
My question now is, how to obtain the Lagrangian (4) directly from the Lagrangian (1). Since a linear EOM is obtained by a quadratic Lagrangian, linearising (1) does not do the trick. One could "quadratise" (1) to get (4), and assume that the variation of the action associated to (1) vanishes, such that the actual Lagrangian for $\varepsilon$ reads
$$\begin{align} \tilde{\mathcal{L}}(\varepsilon, \nabla \varepsilon) &= \frac{1}{2} \, g^{\mu \nu} \, \partial_{\mu} \phi \; \partial_{\nu} \phi + \sum_{k=0}^{2} ( \phi-\phi_0 )^k \cdot \frac{1}{k!} \frac{\partial^{(k)} V}{\partial \phi^k} \, \bigg|_{\phi_0} \\&{} \\&= \mathcal{L}(\phi_0, \nabla \phi_0) + \frac{1}{2} \, g^{\mu \nu} \,\partial_{\mu} \varepsilon \; \partial_{\nu} \varepsilon + \varepsilon \cdot \frac{\partial V}{\partial \phi} \, \bigg|_{\phi_0} + \varepsilon^2 \cdot \frac{1}{2} \frac{\partial^2 V}{\partial \phi^2} \, \bigg|_{\phi_0} \tag{5} \end{align}$$
but then what happens to the term linear in $\varepsilon$? So what actually is "linearising" an equation on the level of the Lagrangian?