On a particular definition of co-variation process. Let $Z$ and $V$ be continuous processes with bounded variation and $X,Y$ be continous local martingales. We have given the following definition in class:
$$[ X+Z,Y+V]:= [X,Y]$$
where $[X,Y]$ is the co-variation process of $X$ and $Y$, defined as
$$[X,Y]_t :=\mathbb{P}- \lim_{n \to \infty} \sum_{j=0}^n (X_{t_j}-X_{t_{j-1}})(Y_{t_j}-Y_{t_{j-1}})$$
This immediately struck me more as a theorem than as a definition but for the sake of simplicity we where using it as a definition, is my intuition true? 
Is the problem the fact that  $[ X+Z,Y+V] $ is not defined for non martingales?
 A: Yes, that's correct, it's actually a theorem (which is not that difficult to prove):
Let's define
$$[X,Y]_t := \lim_n \sum_{j=1}^n (X_{t_j}-X_{t_{j-1}})(Y_{t_j}-Y_{t_{j-1}})$$
whenever the limit exists. Clearly, we have
$$\begin{align*} [X+Z,Y]_t &= [X,Y]_t + [Z,Y]_t  \tag{1} \\ [X,Y]_t &= [Y,X]_t \tag{2} \end{align*}$$
The key is the following lemma:

Let $(X_t)_{t \geq 0}$ be a stochastic process with continuous sample paths and $(Y_t)_{t \geq 0}$ a process of bounded variation. Then $[X,Y]_t = 0$ for all $t$.

Proof: Let $\Pi = \{t_0<\ldots<t_n\}=t$ be a partition of the interval $[0,t]$ for fixed $t>0$. Since $(Y_t)_{t \geq 0}$ is a process of bounded variation, we have $$\left|  \sum_{j=0}^n (X_{t_j}-X_{t_{j-1}})(Y_{t_j}-Y_{t_{j-1}}) \right| \leq \sup_{\substack{u,v \in [0,t] \\ |u-v| \leq |\Pi|}} |X_u-X_v| \underbrace{\sum_{j=1}^n |Y_{t_j}-Y_{t_{j-1}}|}_{\leq V_t(Y)<\infty}$$ where $V_t(Y)$ denotes the variation of $[0,t] \ni s \mapsto Y_s$. Using that $[0,t] \ni s \mapsto X_s(\omega)$ is uniformly continuous, we find that $$\left|  \sum_{j=0}^n (X_{t_j}-X_{t_{j-1}})(Y_{t_j}-Y_{t_{j-1}}) \right|$$ converges (almost surely) to $0$ as the mesh size $|\Pi|$ tends to zero. This finishes the proof.

Now let $Z,V$ be two processes with continuous sample paths of bounded variation and $X,Y$ two stochastic processes with continuous sample paths such that $[X,Y]_t$ exists. Then we find from $(1)$, $(2)$ and the above lemma
$$\begin{align*} [X+Z,Y+V]_t = [X,Y+V]_t + [Z,Y+V]_t &= [X,Y]_t + \underbrace{[X,V]_t}_{0} + \underbrace{[Z,Y]_t}_{0} + \underbrace{[Z,V]_t}_{0} = [X,Y]_t. \end{align*}$$
