Let $G$ be a directed graph with edge set $E$ and $u:E \to \mathbb{R}_{\geq 0}$ a capacity function, as well as $x: E \to \mathbb{R}_{\geq 0}$ be a function. Then $x$ is called an $s-t$ flow if the following properties are satisfied:
$(a)$ $x(e) \leq u(e)$ for all $e \in E$,
$(b)$ for all $v \in V - \{s,t\}$ it holds that: $\sum_{e \in \delta^{+}(v)} x(e)-\sum_{e \in \delta^{-}(v)} x(e)=0,$ where $\delta^+(v)$ denotes the edges that are starting in $v$ and $\delta^{-}(v)$ are the edges that are ending in $v$.
The flow value $v(x)$ is now defined as $v(x):=\sum_{e \in \delta^{+}(s)} x(e)-\sum_{e \in \delta^{-}(s)} x(e)$, or equivalently $v(x):=\sum_{e \in \delta^{+}(t)} x(e)-\sum_{e \in \delta^{-}(t)} x(e)$. My question is, why are those equivalent? My idea was to sum the equation in $(b)$ over all $v \in V - \{s,t\}$, giving $$0= \sum_{v \in V -\{s,t\}}(\sum_{e \in \delta^{+}(v)} x(e)-\sum_{e \in \delta^{-}(v)} x(e)) \iff v(x)=\sum_{v \in V -\{t\}}(\sum_{e \in \delta^{+}(v)} x(e)-\sum_{e \in \delta^{-}(v)} x(e)),$$ where I am using the first definition for $v(x)$. However, I don't know how to proceed, is this last expression equal to the other definition of $v(x)$?
You should sum over all vertices.
For simplicity, I adopt the notation $x(\delta^+(v))=\sum_{e\in \delta^+(v)}x(e)$ and analogously for $\delta^-(v)$ for any $v\in V$. We have
$$\sum_{v\in V}x(\delta^+(v))=\sum_{v\in V}x(\delta^-(v))$$ because every edge is ingoing exactly once and outgoing exactly once (i.e. every edge is counted exactly once on both sides).
Then essentially subtract what you have calculated:
$$\sum_{v\in V}x(\delta^+(v))-\sum_{v\neq s,t}(\delta^+(v))=\sum_{v\in V}x(\delta^-(v))-\sum_{v\neq s,t}(\delta^-(v))$$
This holds because $\sum_{v\neq s,t}(\delta^+(v))=\sum_{v\neq s,t}(\delta^-(v))$ by the second property. But this just means
$$\delta^+(s)+\delta^+(t)=\delta^-(s)+\delta^-(t)$$
which yields the result
$$\delta^+(s)-\delta^-(s)=\delta^-(t)-\delta^+(t)$$
by a simple arrangement.