I will give an elementary proof of the problem using the fact, that $T^2$ is a topological group and that its universal cover is contractible. We will start with some constructions in homotopy theory of topological groups, which are required to understand the proof given below and added here for convenience.
First note, that $\mathbb R^2$ forms a topological group under addition: $(x,y) + (x',y') := (x+x',y+y')$ and that $\mathbb Z^2$ is a discrete normal subgroup thereof. We identify $T^2$ with the quotient of $\mathbb R^2$ by $\mathbb Z^2$, so that $T^2$ again becomes a topological group under addition. Moreover the quotient map $p: \mathbb R^2 \to T^2$ becomes a group homomorphism and is easily seen to be the universal covering projection ($\mathbb Z^2$ discrete subgroup $\Rightarrow$ $p$ is covering projection; $\mathbb R^2$ is contractible $\Rightarrow$ $p$ is universal).
Next, we observe, that for $[f],[g]\in [(X,\ast),(T^2,0)]$ the sum $[f]+[g] := [f+g]$ is well defined, turning $[(X,\ast),(T^2,0)]$ into a group. The same arguments show that $[(X,\ast),(\mathbb R^2,0)]$ is a group under point wise addition of representatives as well (as is $[(X,\ast),(G,1)]$ for any topological group $G$ with unit $1$), and that the map $p_\sharp : [(X,\ast),(\mathbb R^2,0)] \to [(X,\ast),(T^2,0)]$ given by $p_\sharp([f]) := [p \circ f]$ is a group homomorphism.
Now $\pi_1(T^2,0) = [(S^1,1), (T^2,0)]$ is a group in two ways, by means of composition of (representatives of) loops $[\alpha], [\beta] \mapsto [\alpha \ast \beta]$ and by means of point wise addition of (representatives of) loops $[\alpha], [\beta] \mapsto [\alpha + \beta]$.
Both operations share the same unit, the (class of the) constant loop sending everything to $0 \in T^2$ and denoted simply by $0: (S^1,1) \to (T^2,0)$. We can also observe, that for any loops $\alpha, \beta, \gamma, \delta$ we have $(\alpha + \beta) \ast (\gamma + \delta) = (\alpha + \gamma) \ast (\beta + \delta)$. Therefore $$[\alpha]+[\beta] = ([\alpha] \ast [0]) + ([0] \ast [\beta]) = ([\alpha] + [0]) \ast ([0] + [\beta]) = [\alpha] \ast [\beta],$$
hence the two operations are in fact the same on $\pi_1(T_2,0)$. The same argument can be used to show the analogous statement for $\pi_1(\mathbb R^2,0)$ (or $\pi_1(G,1)$ for any topological group $G$ with unit $1$).
Now back to the problem:
Given two maps $\varphi, \psi: T^2 \to T^2$, such that for some point $x \in T^2$, we have $\varphi(x) = \psi(x) = x$ and $\pi_1(\varphi) = \pi_1(\psi): \pi_1(T^2,x) \to \pi_1(T^2,x)$, we want to show $\varphi \simeq \psi$, where the homotopy can be taken relative to $x$. Replacing $\varphi$ with $\xi \mapsto \varphi(\xi + x) - x$ and $\psi$ with $\xi \mapsto \psi(\xi + x) - x$ if necessary, we may assume $x=0$. It will then suffice to show, that $\chi \simeq 0$, where $\chi := \varphi - \psi$.
Since the induced map $\pi_1(\chi): \pi_1(T^2,0) \to \pi_1(T^2,0)$ on fundamental groups is trivial (this is where we need all the constructions for topological groups), we can lift $\chi$ to a map $\bar{\chi}: (T^2,0) \to (\mathbb R^2,0)$ with $\chi = p \circ \bar{\chi}$.
We now define $H: T^2 \times I \to T^2$ by $H(x,t) = p(t\bar\chi(x))$, which is easily checked to be the required homotopy $0 \simeq \chi$.
In this context, when one says $\phi, \psi: (X, x_0)\to(Y, y_0)$ are homotopic, that means they are homotopic as maps $(X,x_0)\to(Y,y_0)$, not just as maps $X\to Y$. This means that every stage of the homotopy preserves the basepoint: $H(x_0,t)=y_0$ for all $t$. This guarantees that $K$ is a path homotopy.
(Without this stronger assumption on the homotopy from $\phi$ to $\psi$, it is not necessarily true that they induce the same map on fundamental groups. In general, they will differ by conjugation by the element of $\pi_1(Y,y_0)$ defined by the loop $t\mapsto H(x_0,t)$.)
Best Answer
This is essentially telling you that if two maps $X\to Y$ are homotopic, then they induce the "same" map at the level of homotopy groups, where the meaning of "same" here is only as broad as permitted by the fact that the two induced maps actually don't have the same codomain strictly speaking, because of the base point.
What could potentially be enlightening is to wonder what would happen if the two maps were not homotopic.
For instance consider these few examples:
$X$ is connected, $Y$ is not, and $\varphi$ and $\psi$ land in two different components of $Y$. Obviously the induced maps in homotopy can't be considered equal in any reasonable way.
$X=\Bbb S^1$ with some base point that you call $1$, $Y=\Bbb S^1\times \Bbb S^1$, $\varphi(x)=(x,1)$ and $\psi(x)=(1,x)$. Here, it is interesting to see that there is an isomorphism $u$ from the codomain to itself such that $\psi_* = u\circ \varphi_*$: indeed, $$\varphi_\star =\Bbb Z\to\Bbb Z^2:n\to (n,0)$$ and $$\psi_\star =\Bbb Z\to\Bbb Z^2:n\to (0,n)$$ so that the isomorphism $u$ is simply $$u=\Bbb Z^2\to\Bbb Z^2:(m,n)\to (n,m)$$ But to find this isomorphism you had to understand what was going on and build it yourself. The theorem is actually telling you that when the two maps are homotopic, you don't have to think, $u$ is God-given.