A simple interpretation of algebraic cycles as homology classes I understand an algebraic cycle to be a formal linear combination of closed irreducible subvarieties of some variety $X$ (over $\mathbb{C}$).
I guess to make things concrete, make $X$ smooth and projective with dimension $n$.
Now I have read that a $\operatorname{codim} p$-cycle $Z$ in $X$ defines a $(2n-2p)$-dimensional homology class $[Z] \in H_{2n-2p}(X)$.
My question is

What/is there a pedestrian explanation of why this is the case, and how to visualise it?

Even an example with say $\mathbb{P}^1$ would be helpful here.
I know very little algebraic geometry overall (say only Hartshorne I and the first few chapters of II), so hopefully a minimal amount of reference to say divisors or intersection theory would be ideal (but I'm not sure if this is possible!)
 A: Question: "What/is there a pedestrian explanation of why this is the case, and how to visualise it?"
Answer: After 19.1.1 in Fulton "Intersection theory" they define the cycle map $cl: A_*(X) \rightarrow H_*(X)$ for any "Borel-Moore homology theory" $H_*$. The group $H_{2n}(X)$ contains the fundamental class $cl(X_i):=[X_i]$ for all its n-dimensional irreducible components $X_i$ and to any closed subvariety $V \subseteq X$ you get a class
$$ cl_X(V):=i_*cl(V) \in H_{2k}(X)$$
where $i_*$ is push forward of "Borel-Moore homology" and $cl(V)\in H_{2k}(V)$.
This defines a map
$$cl(X): Z_k(X) \rightarrow H_{2k}(X)$$
by
$$cl(X)(\sum_i n_i[V_i]):= \sum n_i cl_X(V_i) \in H_{2k}(X).$$
In proposition 19.1.1 in the same book it is shown that if $\alpha \cong 0$ is rationally equivalent to zero it follows $cl(X)(\alpha)=0$, hence you get a well defined map
$$cl: A_k(X) \rightarrow H_{2k}(X)$$
for any such $H_*$
Question: "Even an example with say P1 would be helpful here. I know very little algebraic geometry overall (say only Hartshorne I and the first few chapters of II), so hopefully a minimal amount of reference to say divisors or intersection theory would be ideal (but I'm not sure if this is possible!)"
Answer: If $H_*$ is singular homology and if $X$ is any irreducible n-dimensional complex algebraic variety and $V \subseteq X$ a k-dimensional subvariety we may view $V,X$ as topological spaces in the strong topology and get a push forward map
$$i_*:H_*(V) \rightarrow H_*(X)$$
and since the "fundamental class" $[V]$ of $V$ generates $H_{2k}(V)$ there is a canonical element $i_*[V]\in H_{2k}(X)$ and this defines the cycle map $cl_X([V]):=i_*[V]$. In the case of $C:=\mathbb{P}^1_{\mathbb{C}}$ any divisor  $D:= \sum_i n_i[P_i]$ gives a formal sum where $P_i \in C$ is a closed point and you get an element $i_*[P_i] \in H_*(C)$. This gives rise to  the cycle map
$$cl_C: A_*(C)\rightarrow H_*(C).$$
The difficult part here is to prove that cycles that are rationally equivalent to zero maps to zero under $cl_C$. By the projective bundle formula it follows $A_*(C) \cong\mathbb{Z}[t]/(t^2)$.
If you use the singular chain complex $C_n(X)$ to define $H_n(X):=ker(d_n)/Im(d_{n+1})$, it should be possible to give an "elementary construction" and to prove that for any cycle $\alpha_k \in Z_k(X)$ rationally equivalent to zero, there is a topological cycle $z\in C_{2k+1}(X)$ with $d_{2k+1}(z)=i_*(\alpha_k)$, hence $cl_X(\alpha_k)=0$ in $H_{2k}(X)$. I believe this is possible and has been written down but I do not have a precise reference to the litterature.
