Is the reduction for HAMPATH to HAMCYCLE and UHAMPATH to UHAMCYCLE the same? HAMPATH/UHAMPATH is A directed / undirected graph G and 2 nodes s and t and is there a hamilton path from s to t? 

Likewise with HAMCYCLE/UHAMCYCLE but has a hamilton cycle on $G'$ 

The reduction for directed is

$HAMPATH \leq_p HAMCYCLE$

function(G, s, t)
    G' = G
    add node t' to G'
    add edges (t, t') and (t', s)
    return (G')


$(\Rightarrow)$If $(G, s, t) \in HAMPATH$, then $\{(s, v_1), \dots, (v_n, t)\}$ is a hamilton path in $G$ from $s$ to $t$ and our reduction for $G'$ has a path $\{(s, v_1), \dots, (v_n, t), (t, t'), (t', s)\}$ which is a hamilton cycle $(G') \in HAMCYCLE$

$(\Leftarrow)$ If $(G') \in HAMCYCLE$, then $\{(s, v_1), \dots, (v_n, t), (t, t'), (t', s)\}$ is a hamilton cycle in $G$. Remove the edges $(t, t')$ and $(t', s)$ and we are left with a hamilton path $\{(s, v_1), \dots, (v_n, t)\}$, $(G, s, t) \in HAMPATH$



Wouldn't this reduction also work for the undirected version i.e. $UHAMPATH \leq_p UHAMCYCLE$? I've seen on the internet another reduction where they add another node and add an edge between every vertex in G for $G'$ but wouldn't this simpler version work as well?
Your reduction is correct. However, the standard Hamiltonian path problem does not specify the endnodes $s$ and $t$, which is a bit different from your definition. This is possibly why the internet does not use your simple reduction.