The (simple) Markov property
$$\mathbb{P}(X_t \in A \mid \mathcal{F}_s) = \mathbb{P}(X_t \in A \mid X_s) \tag{1}$$
makes perfect sense in any dimension $n \geq 1$. If, say, $(X_t)_{t \geq 0}$ is a continuous stochastic process which takes values in $\mathbb{R}^n$, then $(1)$ is well-defined for any Borel set $A \in \mathcal{B}(\mathbb{R}^n)$. The interpretation of $(1)$ is the same for any dimension $n \geq 1$: The evolution of the process in the future does only depend on the present state and not on the past. A Markov process is memoryless in the sense that it does not remember the past but only the present.
If some process $(X_t)_{t \geq 0}$ is not Markovian then $Z_t := (X_t,Y_t)_{t \geq 0}$ still might be Markovian. Why? Adding another component $(Y_t)_{t \geq 0}$ means that the new process has a larger information about the present - note that $Z_t = (X_t,Y_t)$ gives clearly more information than just the value of $X_t$.
Let's consider a Brownian motion $(B_t)_{t \geq 0}$ and $X_t := \int_0^t B_r \, dr$. Let's first try to get some intuition why $(X_t)_{t \geq 0}$ is not Markovian. Fix $s \leq t$, i.e. $s$ corresponds to "present" and $t$ is the "future". Clearly,
$$X_t = X_s + \int_s^t B_r \, dr.$$
This tells us the following: The evolution of the process in the future depends on the present state $X_s$ and on $\int_s^t B_r \, dr$. For $(X_t)_{t \geq 0}$ to be Markovian we would need to show that $\int_s^t B_r \, dr$ depends only on the present state $X_s$ and not on the past - but that's impossible since the value of the integral $\int_s^t B_r \, dr$ depends highly on $B_s$ (e.g. if $B_s$ is very large, then $\int_s^t B_r \, dr$ will be large (at least for $t$ close to $s$)), and the present $X_s$ does not give us any information about $B_s$. However, this indicates that we might have a chance to prove that $(X_t,B_t)_{t \geq 0}$ is Markovian.
To prove that $(X_t,B_t)_{t \geq 0}$ is Markovian, we note that
$$X_t = X_s + \int_s^t B_r \, dr = X_s + B_s (t-s) + \int_s^t (B_r-B_s) \, dr$$
and
$$B_t = B_s + (B_t-B_s).$$
Combining both equations we find that
$$\begin{pmatrix} X_t \\ B_t \end{pmatrix} = f \begin{pmatrix} X_s \\ B_s \end{pmatrix} + Z$$
where $f$ is a deterministic function and $Z$ is a suitable random variable which is independent from past and present $(X_r,B_r)_{r \leq s}$ (due to the independence of the increments of Brownian motion). Now our interpretation of $(1)$ tells us that $(X_t,B_t)_{t \geq 0}$ is Markovian: the evolution of $(X_t,B_t)$ in the future does not depend on the past but only on the present $(X_s,B_s)$.
Formally, the proof goes as follows: Denote by $\mathcal{F}_s = \sigma(B_r;r \leq s)$ the canonical filtration of the Brownian motion. Take a bounded Borel measurable function $u:\mathbb{R}^2 \to \mathbb{R}$, then
\begin{align*} \mathbb{E} \left( u(X_t,B_t) \mid \mathcal{F}_s \right) &= \mathbb{E} \left( u(X_s+(t-s)B_s + \int_s^t (B_r-B_s) \, dr, B_s + (B_t-B_s)) \mid \mathcal{F}_s \right). \end{align*}
Since $(B_r-B_s)_{r \geq s}$ is independent from $\mathcal{F}_s$ and $(X_s,B_s)$ is $\mathcal{F}_s$-measurable, it follows that
\begin{align*} \mathbb{E} \left( u(X_t,B_t) \mid \mathcal{F}_s \right) &= g(X_s,B_s) \tag{2} \end{align*}
where
$$g(x,y) := \mathbb{E} \left( u(x + (t-s)y+ \int_s^t (B_r-B_s) \, dr, y + (B_t-B_s)) \right).$$
By the tower property of conditional expectation, $(2)$ implies
\begin{align*} \mathbb{E}(u(X_t,B_t) \mid (X_s,B_s)) &= \mathbb{E} \bigg[ \mathbb{E}(u(X_t,B_t) \mid \mathcal{F}_s) \mid (X_s,B_s)) \bigg] \\ &\stackrel{(2)}{=} g(X_s,B_s). \tag{3} \end{align*}
Combining $(2)$ and $(3)$ we get
\begin{align*} \mathbb{E} \left( u(X_t,B_t) \mid \mathcal{F}_s \right) &= g(X_s,B_s) \\ &= \mathbb{E}(u(X_t,B_t) \mid (X_s,B_s)) \end{align*}
which proves that $(X_t,B_t)_{t \geq 0}$ is Markovian (with respect to $(\mathcal{F}_t)_{t \geq 0}$).
