Consider the functional $$J[y]=\int_{x_1}^{x_2} L(x,y(x),y'(x))\mathrm{d} x.\tag{1}$$ In order to arrive at the classical Euler-Lagrange equation, it is common to build a variation of $y(x)$ in the form $\hat{y}(x,\epsilon)=y(x)+\epsilon v(x)$. However, it might not be needed. Instead, we could just think of an $\epsilon$-parametrized family of functions $\hat{y}(x,\epsilon)$ known to coincide with $y(x)$ when $\epsilon=0$. We also add the usual boundary conditions $\hat{y}(x_1,\epsilon)=\hat{y}(x_2,\epsilon)=0$. Inserted in the above equation, we get: $$J[\hat{y}]=\int_{x_1}^{x_2} L(x,\hat{y}(x,\epsilon),\hat{y}'(x,\epsilon))\mathrm{d} x.\tag{2}$$ Then, enforcing $$\frac{\mathrm{d}J[y]}{\mathrm{d}\epsilon}\Big|_{\epsilon=0}=\int_{x_1}^{x_2}\Bigl( \frac{\partial L}{\partial \hat{y}}\frac{\partial \hat{y}}{\partial \epsilon}+\frac{\partial L}{\partial \hat{y}'}\frac{\partial \hat{y}'}{\partial \epsilon}\Bigr)\Big|_{\epsilon=0}\mathrm{d} x=0.\tag{3}$$ One integration by parts on the second terms leads to: $$\frac{\mathrm{d}J[y]}{\mathrm{d}\epsilon}\Big|_{\epsilon=0}=\int_{x_1}^{x_2}\Bigl(\Bigl( \frac{\partial L}{\partial \hat{y}}-\frac{\mathrm{d}}{\mathrm{d}x}\Bigl(\frac{\partial L}{\partial \hat{y}'}\Bigr)\Bigr)\frac{\partial \hat{y}}{\partial \epsilon}\Bigr)\Big|_{\epsilon=0}\mathrm{d} x=0.\tag{4}$$ We can then invoke the arbitrariness of $\partial \hat{y}/\partial \epsilon$ to obtain the desired result.
The question is thus: why do we always use $\hat{y}(x,\epsilon)=y(x)+\epsilon v(x)$ and how do we know that this is a correct way of parametrizing $\hat{y}(x,\epsilon)$?
As long as $\hat{y}(x,\epsilon)$ is differentiable with respect to $\epsilon$ at $\epsilon=0$, you can write $\hat{y}(x,\epsilon)=\hat{y}(x,0)+\epsilon \partial_\epsilon \hat{y}(x,0) + o(\epsilon)$ and the $o(\epsilon)$ term does not impede the derivation of the Euler-Lagrange equation. Thus the direction of perturbation $v(x)$ is effectively $\partial_\epsilon \hat{y}(x,0)$.