In the context of continuous markov chain,
$P'(t) = P(t)Q$ is called the "forward equation" and
$P'(t) = QP(t)$ is called the "backward equation"
Is this just because in first case Q is in "front" and in the 2nd case Q is in the "back" or is there another explanation for the naming?
The forward equations are obtained by conditioning on the last jump before time $t$ and considering the integral equation for $P(t)$: $$ P_{ij}(t) = e^{-\lambda(i)t}\mathsf 1_{\{i=j\}} + \int_0^t \lambda(i)e^{-\lambda(i)s}\sum_{k\ne i} Q_{ik} P_{kj}(t-s)\ \mathsf ds, $$ where $\lambda(i)$ is the holding time in state $i$ and $Q_{ik}$ is the $(i,k)^{\mathrm{th}}$ entry of the embedded Markov chain. For regular systems, such a last jump exists, but if explosion is possible then the last jump may fail to exist.
Recall the construction of the continuous-time Markov chain $\{X(t):t\geqslant 0\}$: we have a discrete-time Markov chain $\{X_n,n=0,1,2,\ldots\}$ with transition matrix $Q$, an i.i.d. sequence $\{E_n\}$ of unit exponentially distributed random variables independent of $\{X_n\}$, and the holding times $\{\lambda(i):i\in S\}$ with $\lambda(i)>0$ for all $i\in S$ ($S$ being the state space). Let $T_0=0$ and define $W(0) = E_0/\lambda(X_0)$, so that $$ \mathbb P(W(0)>x\mid X_0=i) = e^{-\lambda(i)x}\cdot\mathsf 1_{(0,\infty)}(x). $$ Now define $T_1=T_0+W(0)$, $X(t)\mathsf 1_{\{T_0\leqslant t<T_1\}} = X_0$, and $W(T_1) = E_1/\lambda(X_1)$. Assume inductively that $\{W(T_m):m\leqslant n-1\}$, $\{T_m:0\leqslant m\leqslant n\}$, and $\{X(s): 0\leqslant s<T_n\}$ have been defined. We define $W(T_n) = E_n/\lambda(X_n)$, $T_{n+1} = T_n + W(T_n)$, and $X(t)\mathsf 1_{\{T_n\leqslant t<T_{n+1}\}} = X_n$. Setting $T_\infty=\lim_{n\to\infty} T_n$, we have the process defined on $[0,T_\infty)$, and for $t<T_\infty$, $$ X(t) = \sum_{n=0}^\infty X_n\mathsf 1_{[T_n,T_{n+1})}(t). $$ Now, if $\{E_n\}$ are independent exponentially distributed random variables with $\mathbb P(E_n>x) = e^{-\lambda(n)x}\cdot\mathsf 1_{(0,\infty)}(x)$, then $$ \mathbb P\left(\sum_{n=1}^\infty E_n <\infty\right)=1 \implies \sum_{n=1}^\infty\frac1{\lambda(n)}<\infty. $$ (Proof is left as an exercise to the reader.) Now, for any $i\in S$, $$\mathbb P(T_\infty <\infty\mid X(0)=i) = \mathbb P\left(\sum_{n=1}^\infty \frac1{\lambda(X_n)}<\infty\right) $$ (proof again left as an exercise to the reader).
Therefore the backward equations are more fundamental than the forward equations, because they always have a solution (although the forward equations are often easier to solve).