The standard lifting theorem says that the following:
Let $p : (X,x_0) \to (Y,y_0)$ be a covering map and $Z$ be a path connected locally path connected space. Then $f : (Z,z_0) \to (Y,y_0)$ has a (continuous!) lift $\tilde f : (Z,z_0) \to (X,x_0)$ if and only if $f_*(\pi_1(Z,z_0)) \subset p_*(\pi_1(X,x_0))$.
The condition on fundamental groups is clearly necessary for the existence of a lift (even without any assumptions on $Z$). For the converse let us consider a path connected $Z$. One proceeds as follows:
For each $z \in Z$ choose a path $u : I \to Z$ such that $u(0) = z_0$ and $u(1)= z$. The path $u' = f \circ u : I \to Y$ (which satsfies $u'(0) = y_0$) has a unique lift $\tilde u' : I \to X$ such that $\tilde u'(0) = x_0$. Define
$$\tilde f(z) = \tilde u'(1) .$$
It is easy to see that the condition on fundamental groups assures that $\tilde f(z)$ does not depend on the choice of $u$. Thus we obtain a well-defined function $\tilde f : (Z,z_0) \to (X,x_0)$ such that $p \circ \tilde f = f$.
This works for any path connected $Z$. Let us emphasize that if $f$ has a continuous lift $g = (Z,z_0) \to (X,x_0)$, then it must have necessarily have the form $g = \tilde f$ as above. In fact, given a path $u$ from $z_0$ to $z$, the path $g \circ u$ is a lift of the path $u' = f \circ u$ with $(g \circ u)(0) = g(z_0) = x_0$ and by unique path lifting we conlude $g \circ u = \tilde u'$ and hence $g(z) = (g \circ u)(1) = \tilde u'(1) = \tilde f(z)$.
As you know, the continuity of $\tilde f$ can easily be proved for locally path connected $Z$. But even if $Z$ is not locally path connected, some maps can have continous lifts (for example, all maps with image in an evenly covered set).
Here is an example showing that if $Z$ is not local path connected, then we can get a discontinous $\tilde f$.
Let $p : (\mathbb R,0) \to (S^1,1), p(t) = e^{2\pi i t}$. This is a covering map with extremely nice base and total space. Let $Z$ be the Warsaw circle (see e.g. here and here). This is a simply connected space. Let $L = \{ (0,y):-1\le y \le 1\} \subset Z$ and $q : Z \to Z' = Z/L$ be the quotient map. There exists an obvious homeomorphism $h : S^1 \to Z'$ such $h(-1) = [L]$. Combining it with a rotation on $S^1$, we get a homeomorphism $H : Z' \to S^1$ such that $H([L]) = 1$. Now consider $f = H \circ q : Z \to S^1$ and take $z_0 = (0,0)$. The condition on fundamental groups is satisfied because $Z$ is simply connected, thus we get $\tilde f$ as above. But this map is discontinuous: We have $\tilde f(z_0) = 0$ and $\tilde f(\zeta_n) \to 1$, where the sequence $(\zeta_n)$ has the form $\zeta_n = (t_n,0) \in Z$ with $t_n \to 0$. Look at the graphical representation of $Z$ to get the intuition; there are no "short paths" from $z_0$ to the $\zeta_n$ although they are arbitrarily close to $z_0$.
