From Classical Field Theory by Deligne and Freed, page 143 (page 6 in this pdf) $x$ is a map from $\mathbb{R}$ to $X$ which is Euclidean space with standard inner product $\langle \cdot, \cdot \rangle$. We have Lagrangian density
$$L = \frac{m}{2} |\dot{x}|^2 dt$$
If we deform x we have
$$\delta L = m\langle \dot{x}, \delta \dot{x}\rangle dt = -m\langle \ddot{x},\delta x\rangle dt -d\{m\langle \dot{x}, \delta x\rangle \}$$
It says $\delta$ is the differential on the space of trajectories. What does that mean? What are some resources to get a more basic understanding of what is going on with this manipulation?
You want to take the derivative of $L$ as a function defined on the infinite dimensional space of (sufficiently differentiable) paths in $X$: $\{x:\mathrm R\to X\}$.
This is done by considering a "nearby" path $x+\varepsilon y$ where $y:\mathbb R\to X$ and $\varepsilon \in \mathbb R$ is sent to zero. $$ \begin{align*} \lim_{\varepsilon \to 0}\frac1\varepsilon(L(x+\varepsilon y) - L(x)) &= \lim_{\varepsilon \to 0}\frac1\varepsilon \frac m2 \left[\langle \dot x+\varepsilon \dot y, \dot x+\varepsilon \dot y \rangle - \langle \dot x, \dot x \rangle \right]~dt \\ &= \lim_{\varepsilon \to 0} m\left[ \langle\dot x, \dot y\rangle + \frac{\varepsilon}2\langle \dot y, \dot y\rangle\right] \\ &= m\langle \dot x, \dot y\rangle \end{align*} $$ This will give you the directional derivative of $L$ in the direction $y$, $\mathrm d L(y)$. If we want a "full gradient" of $L$ then we want to rewrite it as an inner product with $y$. Integrate by parts (without the integral) to get $$ \mathrm dL(y) = m\left(\frac{d}{dt} \langle \dot x, y\rangle - \langle \ddot x, y\rangle\right) ~dt $$ If you assume $y$ has compact support, then the boundary term disappears in the integral $\int L~dt$. In that case we see that $\mathrm d L = -m \ddot x$. This differential is also denoted $\delta L$ and is called the variation of $L$ in $x$.
Maybe you can take a look at calculus of variations (wiki)